How to Selection Sort - Assembly 8086





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4















I've created a procedure to selection sort a word vector but there's a problem: the sorting is totally wrong.



My vector: VET_2 DW 2, 7, 0, 1, 4, 8, 9, 3, 6, 5



; Selection Sort
SELECTION_SORT PROC
; AX = j & aux CX = i
; BX = offset/min DX = data and others
PUSH 0 ; to initialize i
MOV SI, [OFFSET VET_2]
; ----- start for(int i = 0; i < n-1; i++) -----
SLC_LOOP_FORA: ; outer loop
CALL RESET_REGIST ; reset some AX, BX, CX & DX
CALL RESET_VAR ; used to reset AUX

POP CX ; initialize i
CMP CX, 18 ; check if it's smaller than n-1 (20-2=18)
JGE SLC_FIM ; if bigger, goes to the end

MOV BX, CX ; offset receive i, the position of the smaller
; ----- start j = i+1 -----
MOV AX, CX ; AX = j.
ADD AX, 2 ; j = i+1
; ----- end j = i+1 -----

; ----- start for(int j = i+1; j < n; j++) -----
SLC_LOOP_DENTRO: ; inner loop
MOV DX, [SI+BX] ; move the smaller element to DX
MOV BX, AX ; offset receives j

CMP DX, [SI+BX] ; compare if VET_2[min]<=VET_2[j]
JL SLC_DENTRO_PULAR ; if lesser, ignore the code below

MOV BX, AX ; offset receive j, position of the smaller element

SLC_DENTRO_PULAR:
ADD AX, 2 ; inc 2 in j
CMP AX, 20 ; compare j (ax) with n
JL SLC_LOOP_DENTRO ; if it's smaller, repeat inner loop
; ----- end for(int j = n+1; j < n; j++) -----

CMP CX, BX ; compare i with the position of the smaller element
JE SLC_LOOP_FORA ; if equals, repeat outer loop, otherwise do the swap

PUSH BX ; position of the smaller element
PUSH [SI+BX] ; put vet[min] top of the stack

; ----- start aux = vet[i] -----
MOV BX, CX ; offset (BX) receives i
MOV DX, [SI+BX] ; DX receives vet_2[i]
MOV AUX, DX ; AUX receives DX
; ----- end aux = vet[i] -----

; ----- start vet[i] = vet[min] -----
POP AX ; AX receives the top of the stack (vet_2[min])
MOV [SI+BX], AX ; vet_2[i] receives DX (smaller element)
; ----- end vet[i] = vet[min] -----

; ----- start vet[min] = aux -----
POP BX ; offset (BX) receives the position of the smaller element from the stack
MOV DX, AUX ; DX receives AUX
MOV [SI+BX], DX ; vet_2[min] receives DX
; ----- end vet[min] = aux -----
ADD CX, 2 ; INC 2 on i
PUSH CX ; put in the stack
JMP SLC_LOOP_FORA repeat outer loop
; ----- end for(int i = 0; i < n-1; i++) -----
SLC_FIM: ; end the procedure
RET
SELECTION_SORT ENDP


Before call selection sort procedure: 2 7 0 1 4 8 9 3 6 5



After call selection sort procedure: 5 2 7 0 1 4 8 9 3 6



Where is the error? Can somebody help me?










share|improve this question




















  • 2





    POP CX inside a loop looks suspicious to me, because you push 0 outside the loop, but there's a JE SLC_LOOP_FORA that can repeat that without pushing anything. So you're going to consume some stack space. xor cx,cx or mov cx,0 would be a lot more sensible. If you run out of registers for locals, normally you store/reload them to locations like [bp-4] after making a stack frame. (You can't use [SP+4] or whatever in 16-bit code). push/pop inside loops is harder to reason about, leading to errors exactly like this.

    – Peter Cordes
    Nov 15 '18 at 16:54






  • 1





    Have you tried single-stepping through your code with a debugger to see if it leaves the loop early? And if so, you can see exactly which branch went wrong.

    – Peter Cordes
    Nov 15 '18 at 17:28


















4















I've created a procedure to selection sort a word vector but there's a problem: the sorting is totally wrong.



My vector: VET_2 DW 2, 7, 0, 1, 4, 8, 9, 3, 6, 5



; Selection Sort
SELECTION_SORT PROC
; AX = j & aux CX = i
; BX = offset/min DX = data and others
PUSH 0 ; to initialize i
MOV SI, [OFFSET VET_2]
; ----- start for(int i = 0; i < n-1; i++) -----
SLC_LOOP_FORA: ; outer loop
CALL RESET_REGIST ; reset some AX, BX, CX & DX
CALL RESET_VAR ; used to reset AUX

POP CX ; initialize i
CMP CX, 18 ; check if it's smaller than n-1 (20-2=18)
JGE SLC_FIM ; if bigger, goes to the end

MOV BX, CX ; offset receive i, the position of the smaller
; ----- start j = i+1 -----
MOV AX, CX ; AX = j.
ADD AX, 2 ; j = i+1
; ----- end j = i+1 -----

; ----- start for(int j = i+1; j < n; j++) -----
SLC_LOOP_DENTRO: ; inner loop
MOV DX, [SI+BX] ; move the smaller element to DX
MOV BX, AX ; offset receives j

CMP DX, [SI+BX] ; compare if VET_2[min]<=VET_2[j]
JL SLC_DENTRO_PULAR ; if lesser, ignore the code below

MOV BX, AX ; offset receive j, position of the smaller element

SLC_DENTRO_PULAR:
ADD AX, 2 ; inc 2 in j
CMP AX, 20 ; compare j (ax) with n
JL SLC_LOOP_DENTRO ; if it's smaller, repeat inner loop
; ----- end for(int j = n+1; j < n; j++) -----

CMP CX, BX ; compare i with the position of the smaller element
JE SLC_LOOP_FORA ; if equals, repeat outer loop, otherwise do the swap

PUSH BX ; position of the smaller element
PUSH [SI+BX] ; put vet[min] top of the stack

; ----- start aux = vet[i] -----
MOV BX, CX ; offset (BX) receives i
MOV DX, [SI+BX] ; DX receives vet_2[i]
MOV AUX, DX ; AUX receives DX
; ----- end aux = vet[i] -----

; ----- start vet[i] = vet[min] -----
POP AX ; AX receives the top of the stack (vet_2[min])
MOV [SI+BX], AX ; vet_2[i] receives DX (smaller element)
; ----- end vet[i] = vet[min] -----

; ----- start vet[min] = aux -----
POP BX ; offset (BX) receives the position of the smaller element from the stack
MOV DX, AUX ; DX receives AUX
MOV [SI+BX], DX ; vet_2[min] receives DX
; ----- end vet[min] = aux -----
ADD CX, 2 ; INC 2 on i
PUSH CX ; put in the stack
JMP SLC_LOOP_FORA repeat outer loop
; ----- end for(int i = 0; i < n-1; i++) -----
SLC_FIM: ; end the procedure
RET
SELECTION_SORT ENDP


Before call selection sort procedure: 2 7 0 1 4 8 9 3 6 5



After call selection sort procedure: 5 2 7 0 1 4 8 9 3 6



Where is the error? Can somebody help me?










share|improve this question




















  • 2





    POP CX inside a loop looks suspicious to me, because you push 0 outside the loop, but there's a JE SLC_LOOP_FORA that can repeat that without pushing anything. So you're going to consume some stack space. xor cx,cx or mov cx,0 would be a lot more sensible. If you run out of registers for locals, normally you store/reload them to locations like [bp-4] after making a stack frame. (You can't use [SP+4] or whatever in 16-bit code). push/pop inside loops is harder to reason about, leading to errors exactly like this.

    – Peter Cordes
    Nov 15 '18 at 16:54






  • 1





    Have you tried single-stepping through your code with a debugger to see if it leaves the loop early? And if so, you can see exactly which branch went wrong.

    – Peter Cordes
    Nov 15 '18 at 17:28














4












4








4








I've created a procedure to selection sort a word vector but there's a problem: the sorting is totally wrong.



My vector: VET_2 DW 2, 7, 0, 1, 4, 8, 9, 3, 6, 5



; Selection Sort
SELECTION_SORT PROC
; AX = j & aux CX = i
; BX = offset/min DX = data and others
PUSH 0 ; to initialize i
MOV SI, [OFFSET VET_2]
; ----- start for(int i = 0; i < n-1; i++) -----
SLC_LOOP_FORA: ; outer loop
CALL RESET_REGIST ; reset some AX, BX, CX & DX
CALL RESET_VAR ; used to reset AUX

POP CX ; initialize i
CMP CX, 18 ; check if it's smaller than n-1 (20-2=18)
JGE SLC_FIM ; if bigger, goes to the end

MOV BX, CX ; offset receive i, the position of the smaller
; ----- start j = i+1 -----
MOV AX, CX ; AX = j.
ADD AX, 2 ; j = i+1
; ----- end j = i+1 -----

; ----- start for(int j = i+1; j < n; j++) -----
SLC_LOOP_DENTRO: ; inner loop
MOV DX, [SI+BX] ; move the smaller element to DX
MOV BX, AX ; offset receives j

CMP DX, [SI+BX] ; compare if VET_2[min]<=VET_2[j]
JL SLC_DENTRO_PULAR ; if lesser, ignore the code below

MOV BX, AX ; offset receive j, position of the smaller element

SLC_DENTRO_PULAR:
ADD AX, 2 ; inc 2 in j
CMP AX, 20 ; compare j (ax) with n
JL SLC_LOOP_DENTRO ; if it's smaller, repeat inner loop
; ----- end for(int j = n+1; j < n; j++) -----

CMP CX, BX ; compare i with the position of the smaller element
JE SLC_LOOP_FORA ; if equals, repeat outer loop, otherwise do the swap

PUSH BX ; position of the smaller element
PUSH [SI+BX] ; put vet[min] top of the stack

; ----- start aux = vet[i] -----
MOV BX, CX ; offset (BX) receives i
MOV DX, [SI+BX] ; DX receives vet_2[i]
MOV AUX, DX ; AUX receives DX
; ----- end aux = vet[i] -----

; ----- start vet[i] = vet[min] -----
POP AX ; AX receives the top of the stack (vet_2[min])
MOV [SI+BX], AX ; vet_2[i] receives DX (smaller element)
; ----- end vet[i] = vet[min] -----

; ----- start vet[min] = aux -----
POP BX ; offset (BX) receives the position of the smaller element from the stack
MOV DX, AUX ; DX receives AUX
MOV [SI+BX], DX ; vet_2[min] receives DX
; ----- end vet[min] = aux -----
ADD CX, 2 ; INC 2 on i
PUSH CX ; put in the stack
JMP SLC_LOOP_FORA repeat outer loop
; ----- end for(int i = 0; i < n-1; i++) -----
SLC_FIM: ; end the procedure
RET
SELECTION_SORT ENDP


Before call selection sort procedure: 2 7 0 1 4 8 9 3 6 5



After call selection sort procedure: 5 2 7 0 1 4 8 9 3 6



Where is the error? Can somebody help me?










share|improve this question
















I've created a procedure to selection sort a word vector but there's a problem: the sorting is totally wrong.



My vector: VET_2 DW 2, 7, 0, 1, 4, 8, 9, 3, 6, 5



; Selection Sort
SELECTION_SORT PROC
; AX = j & aux CX = i
; BX = offset/min DX = data and others
PUSH 0 ; to initialize i
MOV SI, [OFFSET VET_2]
; ----- start for(int i = 0; i < n-1; i++) -----
SLC_LOOP_FORA: ; outer loop
CALL RESET_REGIST ; reset some AX, BX, CX & DX
CALL RESET_VAR ; used to reset AUX

POP CX ; initialize i
CMP CX, 18 ; check if it's smaller than n-1 (20-2=18)
JGE SLC_FIM ; if bigger, goes to the end

MOV BX, CX ; offset receive i, the position of the smaller
; ----- start j = i+1 -----
MOV AX, CX ; AX = j.
ADD AX, 2 ; j = i+1
; ----- end j = i+1 -----

; ----- start for(int j = i+1; j < n; j++) -----
SLC_LOOP_DENTRO: ; inner loop
MOV DX, [SI+BX] ; move the smaller element to DX
MOV BX, AX ; offset receives j

CMP DX, [SI+BX] ; compare if VET_2[min]<=VET_2[j]
JL SLC_DENTRO_PULAR ; if lesser, ignore the code below

MOV BX, AX ; offset receive j, position of the smaller element

SLC_DENTRO_PULAR:
ADD AX, 2 ; inc 2 in j
CMP AX, 20 ; compare j (ax) with n
JL SLC_LOOP_DENTRO ; if it's smaller, repeat inner loop
; ----- end for(int j = n+1; j < n; j++) -----

CMP CX, BX ; compare i with the position of the smaller element
JE SLC_LOOP_FORA ; if equals, repeat outer loop, otherwise do the swap

PUSH BX ; position of the smaller element
PUSH [SI+BX] ; put vet[min] top of the stack

; ----- start aux = vet[i] -----
MOV BX, CX ; offset (BX) receives i
MOV DX, [SI+BX] ; DX receives vet_2[i]
MOV AUX, DX ; AUX receives DX
; ----- end aux = vet[i] -----

; ----- start vet[i] = vet[min] -----
POP AX ; AX receives the top of the stack (vet_2[min])
MOV [SI+BX], AX ; vet_2[i] receives DX (smaller element)
; ----- end vet[i] = vet[min] -----

; ----- start vet[min] = aux -----
POP BX ; offset (BX) receives the position of the smaller element from the stack
MOV DX, AUX ; DX receives AUX
MOV [SI+BX], DX ; vet_2[min] receives DX
; ----- end vet[min] = aux -----
ADD CX, 2 ; INC 2 on i
PUSH CX ; put in the stack
JMP SLC_LOOP_FORA repeat outer loop
; ----- end for(int i = 0; i < n-1; i++) -----
SLC_FIM: ; end the procedure
RET
SELECTION_SORT ENDP


Before call selection sort procedure: 2 7 0 1 4 8 9 3 6 5



After call selection sort procedure: 5 2 7 0 1 4 8 9 3 6



Where is the error? Can somebody help me?







assembly x86-16 selection-sort






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share|improve this question













share|improve this question




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edited Nov 17 '18 at 14:37









Fifoernik

7,85911324




7,85911324










asked Nov 15 '18 at 16:35









Erick MatheusErick Matheus

211




211








  • 2





    POP CX inside a loop looks suspicious to me, because you push 0 outside the loop, but there's a JE SLC_LOOP_FORA that can repeat that without pushing anything. So you're going to consume some stack space. xor cx,cx or mov cx,0 would be a lot more sensible. If you run out of registers for locals, normally you store/reload them to locations like [bp-4] after making a stack frame. (You can't use [SP+4] or whatever in 16-bit code). push/pop inside loops is harder to reason about, leading to errors exactly like this.

    – Peter Cordes
    Nov 15 '18 at 16:54






  • 1





    Have you tried single-stepping through your code with a debugger to see if it leaves the loop early? And if so, you can see exactly which branch went wrong.

    – Peter Cordes
    Nov 15 '18 at 17:28














  • 2





    POP CX inside a loop looks suspicious to me, because you push 0 outside the loop, but there's a JE SLC_LOOP_FORA that can repeat that without pushing anything. So you're going to consume some stack space. xor cx,cx or mov cx,0 would be a lot more sensible. If you run out of registers for locals, normally you store/reload them to locations like [bp-4] after making a stack frame. (You can't use [SP+4] or whatever in 16-bit code). push/pop inside loops is harder to reason about, leading to errors exactly like this.

    – Peter Cordes
    Nov 15 '18 at 16:54






  • 1





    Have you tried single-stepping through your code with a debugger to see if it leaves the loop early? And if so, you can see exactly which branch went wrong.

    – Peter Cordes
    Nov 15 '18 at 17:28








2




2





POP CX inside a loop looks suspicious to me, because you push 0 outside the loop, but there's a JE SLC_LOOP_FORA that can repeat that without pushing anything. So you're going to consume some stack space. xor cx,cx or mov cx,0 would be a lot more sensible. If you run out of registers for locals, normally you store/reload them to locations like [bp-4] after making a stack frame. (You can't use [SP+4] or whatever in 16-bit code). push/pop inside loops is harder to reason about, leading to errors exactly like this.

– Peter Cordes
Nov 15 '18 at 16:54





POP CX inside a loop looks suspicious to me, because you push 0 outside the loop, but there's a JE SLC_LOOP_FORA that can repeat that without pushing anything. So you're going to consume some stack space. xor cx,cx or mov cx,0 would be a lot more sensible. If you run out of registers for locals, normally you store/reload them to locations like [bp-4] after making a stack frame. (You can't use [SP+4] or whatever in 16-bit code). push/pop inside loops is harder to reason about, leading to errors exactly like this.

– Peter Cordes
Nov 15 '18 at 16:54




1




1





Have you tried single-stepping through your code with a debugger to see if it leaves the loop early? And if so, you can see exactly which branch went wrong.

– Peter Cordes
Nov 15 '18 at 17:28





Have you tried single-stepping through your code with a debugger to see if it leaves the loop early? And if so, you can see exactly which branch went wrong.

– Peter Cordes
Nov 15 '18 at 17:28












3 Answers
3






active

oldest

votes


















2














Problem



When you don't need to swap 2 elements, you still need to raise the lower bound in CX. The cmp cx, bx je SLC_LOOP_FORA neglects this. Moreover it leaves you with an unbalanced stack (popping more than was pushed).



Solution:



The problem (included the stack un-balance) is easily corrected by introducing an extra label:



    PUSH 0                 ; to initialize i
MOV SI, OFFSET VET_2
SLC_LOOP_FORA: ; outer loop

...

CMP CX, BX ; compare i with the position of the smaller element
JE NO_SWAP ; if equals, repeat outer loop, otherwise do the swap

...

NO_SWAP:
ADD CX, 2 ; INC 2 on i
PUSH CX ; put in the stack
JMP SLC_LOOP_FORA ; repeat outer loop


Consider




; AX = j & aux      CX = i
; BX = offset/min DX = data and others
PUSH 0 ; to initialize i
MOV SI, [OFFSET VET_2]



If you were willing to re-assign registers, you could enormously simplify this program.



With register assignments like



; AX = j & aux      DI = i
; SI = offset/min DX = data and others
PUSH 0 ; to initialize i
MOV BX, OFFSET VET_2


the swap code e.g. would become just these 3 instructions:



mov  ax, [bx+si]
xchg ax, [bx+di]
mov [bx+si], ax





share|improve this answer































    1















    MOV SI, [OFFSET VET_2]



    I've never seen an assembler that would do it this way! The above sets the SI register equal to the contents of the first array element, so SI=2. Not really useful.



    Acceptable instructions to get the address of your vector are:



    mov si, offset VET_2


    or



    lea si, [VET_2]


    or



    lea si, VET_2





    share|improve this answer



















    • 2





      According to Ross' answer on Confusing brackets in MASM32 mov si, [OFFSET VET_2] is still an immediate. MASM apparently only respects square brackets around numeric literals, or if you use ds:[stuff]. (Seems insane and terrible, and I wouldn't recommend ever using square brackets around an immediate except as intentional obfuscation, but if you're stuck using MASM then that's how it works.)

      – Peter Cordes
      Nov 17 '18 at 14:58













    • +1 for exposing this pure madness in MASM. Thanks to @PeterCordes for the link to Ross' answer. I had no idea...

      – Sep Roland
      Nov 25 '18 at 22:04



















    -3














    selection_sort_i64:
    ; /*
    ; Input:
    ; RDI = long long * array
    ; RSI = length
    ;
    ; Pseudo-C code:
    ;
    ; for (int i = 0; i < n - 1; ++i) {
    ; min = i
    ; for (int j = i + 1; j < n; ++j) {
    ; if a[j] < a[min]; min = j
    ; swap(i, min)
    ;*/

    I64_SIZE equ 8
    LG_I64_SIZE equ 3

    cmp rsi, 1
    jle .end ; Just end it if length <= 1
    xchg rsi, rdi ; rsi => left pointer
    mov rcx, rdi ; rcx => length

    ; RDX will be the boundary for i: RSI + (N-1)*sizeof(int64)
    lea rdx, [rsi + rcx * LL_SIZE - LL_SIZE]

    ; /*
    ; Let's explain what's happening here.
    ; RSI will be &a[i], RDX will be it's right boundary
    ; RDI will be &a[j] (&a[i + 1]) and will loop n-i times
    ; RCX will be n-i and will be the counter for the inner loop
    ; RAX will track our minimum in the remaining of the array
    ; RBX will
    ; */
    .outer_loop:
    cmp rsi, rdx
    jge .end ; while (i < n - 1) {
    mov rax, [rsi] ; min = a[i]
    mov rbx, rsi ; min_i = i
    push rax

    mov rdi, rsi
    add rdi, I64_SIZE ; j = i + 1 // rdi => right pointer
    dec rcx ; // rcx = n - i, inner loop counter
    push rcx
    .inner_loop:
    cmp rax, [rdi] ; if (min > a[j])
    jle .min_less_or_equal
    mov rbx, rdi ; min_i = j
    mov rax, [rdi] ; min = a[j]
    .min_less_or_equal:
    add rdi, I64_SIZE ; j += 1
    loop .inner_loop
    pop rcx ; // restore inner loop counter
    pop rax ; // restore a[i]

    cmp rsi, rbx ; // swap if minimum found
    jz .no_min ; if (i != min_i)
    mov rdi, [rbx] ; temp = a[min]
    mov [rbx], rax ; a[min] = a[i]
    mov [rsi], rdi ; a[i] = temp
    .no_min:
    add rsi, I64_SIZE ; i += 1
    jmp .outer_loop ; } // end outer loop
    .end:
    ret


    Hope this works. Thanks.






    share|improve this answer





















    • 2





      The OP is looking for help debugging their code, not for some random selection-sort implementation. This is an x86-16 question, but at least you didn't use any of r8..r15 or any addressing modes that aren't available in 16-bit mode, so you actually could port it to x86-64 directly just by changing the register sizes and I64_SIZE.

      – Peter Cordes
      Nov 16 '18 at 15:25











    • However, in x86-64 you should be using r8 or something for one of the loop counters, so you don't need to push/pop rcx inside the loop. The loop instruction is slow; only use it if optimizing for code-size over speed. Why is the loop instruction slow? Couldn't Intel have implemented it efficiently?

      – Peter Cordes
      Nov 16 '18 at 15:28














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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    2














    Problem



    When you don't need to swap 2 elements, you still need to raise the lower bound in CX. The cmp cx, bx je SLC_LOOP_FORA neglects this. Moreover it leaves you with an unbalanced stack (popping more than was pushed).



    Solution:



    The problem (included the stack un-balance) is easily corrected by introducing an extra label:



        PUSH 0                 ; to initialize i
    MOV SI, OFFSET VET_2
    SLC_LOOP_FORA: ; outer loop

    ...

    CMP CX, BX ; compare i with the position of the smaller element
    JE NO_SWAP ; if equals, repeat outer loop, otherwise do the swap

    ...

    NO_SWAP:
    ADD CX, 2 ; INC 2 on i
    PUSH CX ; put in the stack
    JMP SLC_LOOP_FORA ; repeat outer loop


    Consider




    ; AX = j & aux      CX = i
    ; BX = offset/min DX = data and others
    PUSH 0 ; to initialize i
    MOV SI, [OFFSET VET_2]



    If you were willing to re-assign registers, you could enormously simplify this program.



    With register assignments like



    ; AX = j & aux      DI = i
    ; SI = offset/min DX = data and others
    PUSH 0 ; to initialize i
    MOV BX, OFFSET VET_2


    the swap code e.g. would become just these 3 instructions:



    mov  ax, [bx+si]
    xchg ax, [bx+di]
    mov [bx+si], ax





    share|improve this answer




























      2














      Problem



      When you don't need to swap 2 elements, you still need to raise the lower bound in CX. The cmp cx, bx je SLC_LOOP_FORA neglects this. Moreover it leaves you with an unbalanced stack (popping more than was pushed).



      Solution:



      The problem (included the stack un-balance) is easily corrected by introducing an extra label:



          PUSH 0                 ; to initialize i
      MOV SI, OFFSET VET_2
      SLC_LOOP_FORA: ; outer loop

      ...

      CMP CX, BX ; compare i with the position of the smaller element
      JE NO_SWAP ; if equals, repeat outer loop, otherwise do the swap

      ...

      NO_SWAP:
      ADD CX, 2 ; INC 2 on i
      PUSH CX ; put in the stack
      JMP SLC_LOOP_FORA ; repeat outer loop


      Consider




      ; AX = j & aux      CX = i
      ; BX = offset/min DX = data and others
      PUSH 0 ; to initialize i
      MOV SI, [OFFSET VET_2]



      If you were willing to re-assign registers, you could enormously simplify this program.



      With register assignments like



      ; AX = j & aux      DI = i
      ; SI = offset/min DX = data and others
      PUSH 0 ; to initialize i
      MOV BX, OFFSET VET_2


      the swap code e.g. would become just these 3 instructions:



      mov  ax, [bx+si]
      xchg ax, [bx+di]
      mov [bx+si], ax





      share|improve this answer


























        2












        2








        2







        Problem



        When you don't need to swap 2 elements, you still need to raise the lower bound in CX. The cmp cx, bx je SLC_LOOP_FORA neglects this. Moreover it leaves you with an unbalanced stack (popping more than was pushed).



        Solution:



        The problem (included the stack un-balance) is easily corrected by introducing an extra label:



            PUSH 0                 ; to initialize i
        MOV SI, OFFSET VET_2
        SLC_LOOP_FORA: ; outer loop

        ...

        CMP CX, BX ; compare i with the position of the smaller element
        JE NO_SWAP ; if equals, repeat outer loop, otherwise do the swap

        ...

        NO_SWAP:
        ADD CX, 2 ; INC 2 on i
        PUSH CX ; put in the stack
        JMP SLC_LOOP_FORA ; repeat outer loop


        Consider




        ; AX = j & aux      CX = i
        ; BX = offset/min DX = data and others
        PUSH 0 ; to initialize i
        MOV SI, [OFFSET VET_2]



        If you were willing to re-assign registers, you could enormously simplify this program.



        With register assignments like



        ; AX = j & aux      DI = i
        ; SI = offset/min DX = data and others
        PUSH 0 ; to initialize i
        MOV BX, OFFSET VET_2


        the swap code e.g. would become just these 3 instructions:



        mov  ax, [bx+si]
        xchg ax, [bx+di]
        mov [bx+si], ax





        share|improve this answer













        Problem



        When you don't need to swap 2 elements, you still need to raise the lower bound in CX. The cmp cx, bx je SLC_LOOP_FORA neglects this. Moreover it leaves you with an unbalanced stack (popping more than was pushed).



        Solution:



        The problem (included the stack un-balance) is easily corrected by introducing an extra label:



            PUSH 0                 ; to initialize i
        MOV SI, OFFSET VET_2
        SLC_LOOP_FORA: ; outer loop

        ...

        CMP CX, BX ; compare i with the position of the smaller element
        JE NO_SWAP ; if equals, repeat outer loop, otherwise do the swap

        ...

        NO_SWAP:
        ADD CX, 2 ; INC 2 on i
        PUSH CX ; put in the stack
        JMP SLC_LOOP_FORA ; repeat outer loop


        Consider




        ; AX = j & aux      CX = i
        ; BX = offset/min DX = data and others
        PUSH 0 ; to initialize i
        MOV SI, [OFFSET VET_2]



        If you were willing to re-assign registers, you could enormously simplify this program.



        With register assignments like



        ; AX = j & aux      DI = i
        ; SI = offset/min DX = data and others
        PUSH 0 ; to initialize i
        MOV BX, OFFSET VET_2


        the swap code e.g. would become just these 3 instructions:



        mov  ax, [bx+si]
        xchg ax, [bx+di]
        mov [bx+si], ax






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 25 '18 at 22:51









        Sep RolandSep Roland

        12.5k22047




        12.5k22047

























            1















            MOV SI, [OFFSET VET_2]



            I've never seen an assembler that would do it this way! The above sets the SI register equal to the contents of the first array element, so SI=2. Not really useful.



            Acceptable instructions to get the address of your vector are:



            mov si, offset VET_2


            or



            lea si, [VET_2]


            or



            lea si, VET_2





            share|improve this answer



















            • 2





              According to Ross' answer on Confusing brackets in MASM32 mov si, [OFFSET VET_2] is still an immediate. MASM apparently only respects square brackets around numeric literals, or if you use ds:[stuff]. (Seems insane and terrible, and I wouldn't recommend ever using square brackets around an immediate except as intentional obfuscation, but if you're stuck using MASM then that's how it works.)

              – Peter Cordes
              Nov 17 '18 at 14:58













            • +1 for exposing this pure madness in MASM. Thanks to @PeterCordes for the link to Ross' answer. I had no idea...

              – Sep Roland
              Nov 25 '18 at 22:04
















            1















            MOV SI, [OFFSET VET_2]



            I've never seen an assembler that would do it this way! The above sets the SI register equal to the contents of the first array element, so SI=2. Not really useful.



            Acceptable instructions to get the address of your vector are:



            mov si, offset VET_2


            or



            lea si, [VET_2]


            or



            lea si, VET_2





            share|improve this answer



















            • 2





              According to Ross' answer on Confusing brackets in MASM32 mov si, [OFFSET VET_2] is still an immediate. MASM apparently only respects square brackets around numeric literals, or if you use ds:[stuff]. (Seems insane and terrible, and I wouldn't recommend ever using square brackets around an immediate except as intentional obfuscation, but if you're stuck using MASM then that's how it works.)

              – Peter Cordes
              Nov 17 '18 at 14:58













            • +1 for exposing this pure madness in MASM. Thanks to @PeterCordes for the link to Ross' answer. I had no idea...

              – Sep Roland
              Nov 25 '18 at 22:04














            1












            1








            1








            MOV SI, [OFFSET VET_2]



            I've never seen an assembler that would do it this way! The above sets the SI register equal to the contents of the first array element, so SI=2. Not really useful.



            Acceptable instructions to get the address of your vector are:



            mov si, offset VET_2


            or



            lea si, [VET_2]


            or



            lea si, VET_2





            share|improve this answer














            MOV SI, [OFFSET VET_2]



            I've never seen an assembler that would do it this way! The above sets the SI register equal to the contents of the first array element, so SI=2. Not really useful.



            Acceptable instructions to get the address of your vector are:



            mov si, offset VET_2


            or



            lea si, [VET_2]


            or



            lea si, VET_2






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 17 '18 at 14:47









            FifoernikFifoernik

            7,85911324




            7,85911324








            • 2





              According to Ross' answer on Confusing brackets in MASM32 mov si, [OFFSET VET_2] is still an immediate. MASM apparently only respects square brackets around numeric literals, or if you use ds:[stuff]. (Seems insane and terrible, and I wouldn't recommend ever using square brackets around an immediate except as intentional obfuscation, but if you're stuck using MASM then that's how it works.)

              – Peter Cordes
              Nov 17 '18 at 14:58













            • +1 for exposing this pure madness in MASM. Thanks to @PeterCordes for the link to Ross' answer. I had no idea...

              – Sep Roland
              Nov 25 '18 at 22:04














            • 2





              According to Ross' answer on Confusing brackets in MASM32 mov si, [OFFSET VET_2] is still an immediate. MASM apparently only respects square brackets around numeric literals, or if you use ds:[stuff]. (Seems insane and terrible, and I wouldn't recommend ever using square brackets around an immediate except as intentional obfuscation, but if you're stuck using MASM then that's how it works.)

              – Peter Cordes
              Nov 17 '18 at 14:58













            • +1 for exposing this pure madness in MASM. Thanks to @PeterCordes for the link to Ross' answer. I had no idea...

              – Sep Roland
              Nov 25 '18 at 22:04








            2




            2





            According to Ross' answer on Confusing brackets in MASM32 mov si, [OFFSET VET_2] is still an immediate. MASM apparently only respects square brackets around numeric literals, or if you use ds:[stuff]. (Seems insane and terrible, and I wouldn't recommend ever using square brackets around an immediate except as intentional obfuscation, but if you're stuck using MASM then that's how it works.)

            – Peter Cordes
            Nov 17 '18 at 14:58







            According to Ross' answer on Confusing brackets in MASM32 mov si, [OFFSET VET_2] is still an immediate. MASM apparently only respects square brackets around numeric literals, or if you use ds:[stuff]. (Seems insane and terrible, and I wouldn't recommend ever using square brackets around an immediate except as intentional obfuscation, but if you're stuck using MASM then that's how it works.)

            – Peter Cordes
            Nov 17 '18 at 14:58















            +1 for exposing this pure madness in MASM. Thanks to @PeterCordes for the link to Ross' answer. I had no idea...

            – Sep Roland
            Nov 25 '18 at 22:04





            +1 for exposing this pure madness in MASM. Thanks to @PeterCordes for the link to Ross' answer. I had no idea...

            – Sep Roland
            Nov 25 '18 at 22:04











            -3














            selection_sort_i64:
            ; /*
            ; Input:
            ; RDI = long long * array
            ; RSI = length
            ;
            ; Pseudo-C code:
            ;
            ; for (int i = 0; i < n - 1; ++i) {
            ; min = i
            ; for (int j = i + 1; j < n; ++j) {
            ; if a[j] < a[min]; min = j
            ; swap(i, min)
            ;*/

            I64_SIZE equ 8
            LG_I64_SIZE equ 3

            cmp rsi, 1
            jle .end ; Just end it if length <= 1
            xchg rsi, rdi ; rsi => left pointer
            mov rcx, rdi ; rcx => length

            ; RDX will be the boundary for i: RSI + (N-1)*sizeof(int64)
            lea rdx, [rsi + rcx * LL_SIZE - LL_SIZE]

            ; /*
            ; Let's explain what's happening here.
            ; RSI will be &a[i], RDX will be it's right boundary
            ; RDI will be &a[j] (&a[i + 1]) and will loop n-i times
            ; RCX will be n-i and will be the counter for the inner loop
            ; RAX will track our minimum in the remaining of the array
            ; RBX will
            ; */
            .outer_loop:
            cmp rsi, rdx
            jge .end ; while (i < n - 1) {
            mov rax, [rsi] ; min = a[i]
            mov rbx, rsi ; min_i = i
            push rax

            mov rdi, rsi
            add rdi, I64_SIZE ; j = i + 1 // rdi => right pointer
            dec rcx ; // rcx = n - i, inner loop counter
            push rcx
            .inner_loop:
            cmp rax, [rdi] ; if (min > a[j])
            jle .min_less_or_equal
            mov rbx, rdi ; min_i = j
            mov rax, [rdi] ; min = a[j]
            .min_less_or_equal:
            add rdi, I64_SIZE ; j += 1
            loop .inner_loop
            pop rcx ; // restore inner loop counter
            pop rax ; // restore a[i]

            cmp rsi, rbx ; // swap if minimum found
            jz .no_min ; if (i != min_i)
            mov rdi, [rbx] ; temp = a[min]
            mov [rbx], rax ; a[min] = a[i]
            mov [rsi], rdi ; a[i] = temp
            .no_min:
            add rsi, I64_SIZE ; i += 1
            jmp .outer_loop ; } // end outer loop
            .end:
            ret


            Hope this works. Thanks.






            share|improve this answer





















            • 2





              The OP is looking for help debugging their code, not for some random selection-sort implementation. This is an x86-16 question, but at least you didn't use any of r8..r15 or any addressing modes that aren't available in 16-bit mode, so you actually could port it to x86-64 directly just by changing the register sizes and I64_SIZE.

              – Peter Cordes
              Nov 16 '18 at 15:25











            • However, in x86-64 you should be using r8 or something for one of the loop counters, so you don't need to push/pop rcx inside the loop. The loop instruction is slow; only use it if optimizing for code-size over speed. Why is the loop instruction slow? Couldn't Intel have implemented it efficiently?

              – Peter Cordes
              Nov 16 '18 at 15:28


















            -3














            selection_sort_i64:
            ; /*
            ; Input:
            ; RDI = long long * array
            ; RSI = length
            ;
            ; Pseudo-C code:
            ;
            ; for (int i = 0; i < n - 1; ++i) {
            ; min = i
            ; for (int j = i + 1; j < n; ++j) {
            ; if a[j] < a[min]; min = j
            ; swap(i, min)
            ;*/

            I64_SIZE equ 8
            LG_I64_SIZE equ 3

            cmp rsi, 1
            jle .end ; Just end it if length <= 1
            xchg rsi, rdi ; rsi => left pointer
            mov rcx, rdi ; rcx => length

            ; RDX will be the boundary for i: RSI + (N-1)*sizeof(int64)
            lea rdx, [rsi + rcx * LL_SIZE - LL_SIZE]

            ; /*
            ; Let's explain what's happening here.
            ; RSI will be &a[i], RDX will be it's right boundary
            ; RDI will be &a[j] (&a[i + 1]) and will loop n-i times
            ; RCX will be n-i and will be the counter for the inner loop
            ; RAX will track our minimum in the remaining of the array
            ; RBX will
            ; */
            .outer_loop:
            cmp rsi, rdx
            jge .end ; while (i < n - 1) {
            mov rax, [rsi] ; min = a[i]
            mov rbx, rsi ; min_i = i
            push rax

            mov rdi, rsi
            add rdi, I64_SIZE ; j = i + 1 // rdi => right pointer
            dec rcx ; // rcx = n - i, inner loop counter
            push rcx
            .inner_loop:
            cmp rax, [rdi] ; if (min > a[j])
            jle .min_less_or_equal
            mov rbx, rdi ; min_i = j
            mov rax, [rdi] ; min = a[j]
            .min_less_or_equal:
            add rdi, I64_SIZE ; j += 1
            loop .inner_loop
            pop rcx ; // restore inner loop counter
            pop rax ; // restore a[i]

            cmp rsi, rbx ; // swap if minimum found
            jz .no_min ; if (i != min_i)
            mov rdi, [rbx] ; temp = a[min]
            mov [rbx], rax ; a[min] = a[i]
            mov [rsi], rdi ; a[i] = temp
            .no_min:
            add rsi, I64_SIZE ; i += 1
            jmp .outer_loop ; } // end outer loop
            .end:
            ret


            Hope this works. Thanks.






            share|improve this answer





















            • 2





              The OP is looking for help debugging their code, not for some random selection-sort implementation. This is an x86-16 question, but at least you didn't use any of r8..r15 or any addressing modes that aren't available in 16-bit mode, so you actually could port it to x86-64 directly just by changing the register sizes and I64_SIZE.

              – Peter Cordes
              Nov 16 '18 at 15:25











            • However, in x86-64 you should be using r8 or something for one of the loop counters, so you don't need to push/pop rcx inside the loop. The loop instruction is slow; only use it if optimizing for code-size over speed. Why is the loop instruction slow? Couldn't Intel have implemented it efficiently?

              – Peter Cordes
              Nov 16 '18 at 15:28
















            -3












            -3








            -3







            selection_sort_i64:
            ; /*
            ; Input:
            ; RDI = long long * array
            ; RSI = length
            ;
            ; Pseudo-C code:
            ;
            ; for (int i = 0; i < n - 1; ++i) {
            ; min = i
            ; for (int j = i + 1; j < n; ++j) {
            ; if a[j] < a[min]; min = j
            ; swap(i, min)
            ;*/

            I64_SIZE equ 8
            LG_I64_SIZE equ 3

            cmp rsi, 1
            jle .end ; Just end it if length <= 1
            xchg rsi, rdi ; rsi => left pointer
            mov rcx, rdi ; rcx => length

            ; RDX will be the boundary for i: RSI + (N-1)*sizeof(int64)
            lea rdx, [rsi + rcx * LL_SIZE - LL_SIZE]

            ; /*
            ; Let's explain what's happening here.
            ; RSI will be &a[i], RDX will be it's right boundary
            ; RDI will be &a[j] (&a[i + 1]) and will loop n-i times
            ; RCX will be n-i and will be the counter for the inner loop
            ; RAX will track our minimum in the remaining of the array
            ; RBX will
            ; */
            .outer_loop:
            cmp rsi, rdx
            jge .end ; while (i < n - 1) {
            mov rax, [rsi] ; min = a[i]
            mov rbx, rsi ; min_i = i
            push rax

            mov rdi, rsi
            add rdi, I64_SIZE ; j = i + 1 // rdi => right pointer
            dec rcx ; // rcx = n - i, inner loop counter
            push rcx
            .inner_loop:
            cmp rax, [rdi] ; if (min > a[j])
            jle .min_less_or_equal
            mov rbx, rdi ; min_i = j
            mov rax, [rdi] ; min = a[j]
            .min_less_or_equal:
            add rdi, I64_SIZE ; j += 1
            loop .inner_loop
            pop rcx ; // restore inner loop counter
            pop rax ; // restore a[i]

            cmp rsi, rbx ; // swap if minimum found
            jz .no_min ; if (i != min_i)
            mov rdi, [rbx] ; temp = a[min]
            mov [rbx], rax ; a[min] = a[i]
            mov [rsi], rdi ; a[i] = temp
            .no_min:
            add rsi, I64_SIZE ; i += 1
            jmp .outer_loop ; } // end outer loop
            .end:
            ret


            Hope this works. Thanks.






            share|improve this answer















            selection_sort_i64:
            ; /*
            ; Input:
            ; RDI = long long * array
            ; RSI = length
            ;
            ; Pseudo-C code:
            ;
            ; for (int i = 0; i < n - 1; ++i) {
            ; min = i
            ; for (int j = i + 1; j < n; ++j) {
            ; if a[j] < a[min]; min = j
            ; swap(i, min)
            ;*/

            I64_SIZE equ 8
            LG_I64_SIZE equ 3

            cmp rsi, 1
            jle .end ; Just end it if length <= 1
            xchg rsi, rdi ; rsi => left pointer
            mov rcx, rdi ; rcx => length

            ; RDX will be the boundary for i: RSI + (N-1)*sizeof(int64)
            lea rdx, [rsi + rcx * LL_SIZE - LL_SIZE]

            ; /*
            ; Let's explain what's happening here.
            ; RSI will be &a[i], RDX will be it's right boundary
            ; RDI will be &a[j] (&a[i + 1]) and will loop n-i times
            ; RCX will be n-i and will be the counter for the inner loop
            ; RAX will track our minimum in the remaining of the array
            ; RBX will
            ; */
            .outer_loop:
            cmp rsi, rdx
            jge .end ; while (i < n - 1) {
            mov rax, [rsi] ; min = a[i]
            mov rbx, rsi ; min_i = i
            push rax

            mov rdi, rsi
            add rdi, I64_SIZE ; j = i + 1 // rdi => right pointer
            dec rcx ; // rcx = n - i, inner loop counter
            push rcx
            .inner_loop:
            cmp rax, [rdi] ; if (min > a[j])
            jle .min_less_or_equal
            mov rbx, rdi ; min_i = j
            mov rax, [rdi] ; min = a[j]
            .min_less_or_equal:
            add rdi, I64_SIZE ; j += 1
            loop .inner_loop
            pop rcx ; // restore inner loop counter
            pop rax ; // restore a[i]

            cmp rsi, rbx ; // swap if minimum found
            jz .no_min ; if (i != min_i)
            mov rdi, [rbx] ; temp = a[min]
            mov [rbx], rax ; a[min] = a[i]
            mov [rsi], rdi ; a[i] = temp
            .no_min:
            add rsi, I64_SIZE ; i += 1
            jmp .outer_loop ; } // end outer loop
            .end:
            ret


            Hope this works. Thanks.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 16 '18 at 15:23









            Peter Cordes

            134k19204344




            134k19204344










            answered Nov 16 '18 at 15:08









            Niloy NiilNiloy Niil

            11




            11








            • 2





              The OP is looking for help debugging their code, not for some random selection-sort implementation. This is an x86-16 question, but at least you didn't use any of r8..r15 or any addressing modes that aren't available in 16-bit mode, so you actually could port it to x86-64 directly just by changing the register sizes and I64_SIZE.

              – Peter Cordes
              Nov 16 '18 at 15:25











            • However, in x86-64 you should be using r8 or something for one of the loop counters, so you don't need to push/pop rcx inside the loop. The loop instruction is slow; only use it if optimizing for code-size over speed. Why is the loop instruction slow? Couldn't Intel have implemented it efficiently?

              – Peter Cordes
              Nov 16 '18 at 15:28
















            • 2





              The OP is looking for help debugging their code, not for some random selection-sort implementation. This is an x86-16 question, but at least you didn't use any of r8..r15 or any addressing modes that aren't available in 16-bit mode, so you actually could port it to x86-64 directly just by changing the register sizes and I64_SIZE.

              – Peter Cordes
              Nov 16 '18 at 15:25











            • However, in x86-64 you should be using r8 or something for one of the loop counters, so you don't need to push/pop rcx inside the loop. The loop instruction is slow; only use it if optimizing for code-size over speed. Why is the loop instruction slow? Couldn't Intel have implemented it efficiently?

              – Peter Cordes
              Nov 16 '18 at 15:28










            2




            2





            The OP is looking for help debugging their code, not for some random selection-sort implementation. This is an x86-16 question, but at least you didn't use any of r8..r15 or any addressing modes that aren't available in 16-bit mode, so you actually could port it to x86-64 directly just by changing the register sizes and I64_SIZE.

            – Peter Cordes
            Nov 16 '18 at 15:25





            The OP is looking for help debugging their code, not for some random selection-sort implementation. This is an x86-16 question, but at least you didn't use any of r8..r15 or any addressing modes that aren't available in 16-bit mode, so you actually could port it to x86-64 directly just by changing the register sizes and I64_SIZE.

            – Peter Cordes
            Nov 16 '18 at 15:25













            However, in x86-64 you should be using r8 or something for one of the loop counters, so you don't need to push/pop rcx inside the loop. The loop instruction is slow; only use it if optimizing for code-size over speed. Why is the loop instruction slow? Couldn't Intel have implemented it efficiently?

            – Peter Cordes
            Nov 16 '18 at 15:28







            However, in x86-64 you should be using r8 or something for one of the loop counters, so you don't need to push/pop rcx inside the loop. The loop instruction is slow; only use it if optimizing for code-size over speed. Why is the loop instruction slow? Couldn't Intel have implemented it efficiently?

            – Peter Cordes
            Nov 16 '18 at 15:28




















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