Behaviour of getchar() in a while loop
I am running this c program in the terminal
#include <stdio.h>
int main() {
int result = 0;
while(result <= 0)
{
int result = (getchar() != EOF);
result = 2;
printf("x");
}
printf("outn");
}
After that I type in the word "hello" followed by a return. The result is that I get multiple 'x' characters.
Why doesn't this terminate after the first 'x'?
c getchar
edited Nov 9 '18 at 16:13
chux
83.6k872152
asked Nov 9 '18 at 15:21
theo_vvvtheo_vvv
233
add a comment |
I am running this c program in the terminal
#include <stdio.h>
int main() {
int result = 0;
while(result <= 0)
{
int result = (getchar() != EOF);
result = 2;
printf("x");
}
printf("outn");
}
After that I type in the word "hello" followed by a return. The result is that I get multiple 'x' characters.
Why doesn't this terminate after the first 'x'?
c getchar
edited Nov 9 '18 at 16:13
chux
83.6k872152
asked Nov 9 '18 at 15:21
theo_vvvtheo_vvv
233
1
Tip: a well enabled compiler may report "warning: variable 'result' set but not used [-Wunused-but-set-variable]". Save time and enable all compiler warnings.
– chux
Nov 9 '18 at 16:12
add a comment |
I am running this c program in the terminal
#include <stdio.h>
int main() {
int result = 0;
while(result <= 0)
{
int result = (getchar() != EOF);
result = 2;
printf("x");
}
printf("outn");
}
After that I type in the word "hello" followed by a return. The result is that I get multiple 'x' characters.
Why doesn't this terminate after the first 'x'?
c getchar
edited Nov 9 '18 at 16:13
chux
83.6k872152
asked Nov 9 '18 at 15:21
theo_vvvtheo_vvv
233
I am running this c program in the terminal
#include <stdio.h>
int main() {
int result = 0;
while(result <= 0)
{
int result = (getchar() != EOF);
result = 2;
printf("x");
}
printf("outn");
}
After that I type in the word "hello" followed by a return. The result is that I get multiple 'x' characters.
Why doesn't this terminate after the first 'x'?
c getchar
c getchar
edited Nov 9 '18 at 16:13
chux
83.6k872152
asked Nov 9 '18 at 15:21
theo_vvvtheo_vvv
233
edited Nov 9 '18 at 16:13
chux
83.6k872152
asked Nov 9 '18 at 15:21
theo_vvvtheo_vvv
233
edited Nov 9 '18 at 16:13
chux
83.6k872152
edited Nov 9 '18 at 16:13
chux
83.6k872152
edited Nov 9 '18 at 16:13
chux
83.6k872152
83.6k872152
asked Nov 9 '18 at 15:21
theo_vvvtheo_vvv
233
asked Nov 9 '18 at 15:21
theo_vvvtheo_vvv
233
asked Nov 9 '18 at 15:21
theo_vvvtheo_vvv
233
233
1
Tip: a well enabled compiler may report "warning: variable 'result' set but not used [-Wunused-but-set-variable]". Save time and enable all compiler warnings.
– chux
Nov 9 '18 at 16:12
add a comment |
1
Tip: a well enabled compiler may report "warning: variable 'result' set but not used [-Wunused-but-set-variable]". Save time and enable all compiler warnings.
– chux
Nov 9 '18 at 16:12
1
1
Tip: a well enabled compiler may report "warning: variable 'result' set but not used [-Wunused-but-set-variable]". Save time and enable all compiler warnings.
– chux
Nov 9 '18 at 16:12
Tip: a well enabled compiler may report "warning: variable 'result' set but not used [-Wunused-but-set-variable]". Save time and enable all compiler warnings.
– chux
Nov 9 '18 at 16:12
add a comment |
2 Answers
2
active
oldest
votes
You're re-declaring (shadowing result) inside the while loop. The result that is used in while(result <= 0) is the one that is declared outside the loop.
answered Nov 9 '18 at 15:25
ODYN-KonODYN-Kon
2,4752825
Thanks. A bit stupid, I was kind off blind for the obvious I guess.
– theo_vvv
Nov 9 '18 at 15:29
1
so to address the wording of the question title directly - the mystery had nothing to do with getchar()
– Kevin Olree
Nov 9 '18 at 17:30
add a comment |
Well,
#include <stdio.h>
int main() {
int result = 0; /* here *OUTER* result gets the value 0 */
while(result <= 0) /* THIS MAKES THE While to execute forever */
{
int result = (getchar() != EOF); /* THIS VARIABLE IS ***NOT*** THE outside result variable */
result = 2; /* external block result is not visible here so this assign goes to the above inner result */
printf("x");
/* INNER result CEASES TO EXIST HERE */
}
printf("outn");
}
As you can deduct from the comments, the result variable that is compared in the while test is the outer one, while the inner one hides the outer one, no assignations can be made to it in the body of the loop, so the loop runs forever. You get an infinite string of xs printed on stdout.
answered Nov 15 '18 at 9:41
Luis ColoradoLuis Colorado
4,3601718
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You're re-declaring (shadowing result) inside the while loop. The result that is used in while(result <= 0) is the one that is declared outside the loop.
answered Nov 9 '18 at 15:25
ODYN-KonODYN-Kon
2,4752825
Thanks. A bit stupid, I was kind off blind for the obvious I guess.
– theo_vvv
Nov 9 '18 at 15:29
1
so to address the wording of the question title directly - the mystery had nothing to do with getchar()
– Kevin Olree
Nov 9 '18 at 17:30
add a comment |
You're re-declaring (shadowing result) inside the while loop. The result that is used in while(result <= 0) is the one that is declared outside the loop.
answered Nov 9 '18 at 15:25
ODYN-KonODYN-Kon
2,4752825
Thanks. A bit stupid, I was kind off blind for the obvious I guess.
– theo_vvv
Nov 9 '18 at 15:29
1
so to address the wording of the question title directly - the mystery had nothing to do with getchar()
– Kevin Olree
Nov 9 '18 at 17:30
add a comment |
You're re-declaring (shadowing result) inside the while loop. The result that is used in while(result <= 0) is the one that is declared outside the loop.
answered Nov 9 '18 at 15:25
ODYN-KonODYN-Kon
2,4752825
You're re-declaring (shadowing result) inside the while loop. The result that is used in while(result <= 0) is the one that is declared outside the loop.
answered Nov 9 '18 at 15:25
ODYN-KonODYN-Kon
2,4752825
answered Nov 9 '18 at 15:25
ODYN-KonODYN-Kon
2,4752825
answered Nov 9 '18 at 15:25
ODYN-KonODYN-Kon
2,4752825
answered Nov 9 '18 at 15:25
ODYN-KonODYN-Kon
2,4752825
2,4752825
Thanks. A bit stupid, I was kind off blind for the obvious I guess.
– theo_vvv
Nov 9 '18 at 15:29
1
so to address the wording of the question title directly - the mystery had nothing to do with getchar()
– Kevin Olree
Nov 9 '18 at 17:30
add a comment |
Thanks. A bit stupid, I was kind off blind for the obvious I guess.
– theo_vvv
Nov 9 '18 at 15:29
1
so to address the wording of the question title directly - the mystery had nothing to do with getchar()
– Kevin Olree
Nov 9 '18 at 17:30
Thanks. A bit stupid, I was kind off blind for the obvious I guess.
– theo_vvv
Nov 9 '18 at 15:29
Thanks. A bit stupid, I was kind off blind for the obvious I guess.
– theo_vvv
Nov 9 '18 at 15:29
1
1
so to address the wording of the question title directly - the mystery had nothing to do with getchar()
– Kevin Olree
Nov 9 '18 at 17:30
so to address the wording of the question title directly - the mystery had nothing to do with getchar()
– Kevin Olree
Nov 9 '18 at 17:30
add a comment |
Well,
#include <stdio.h>
int main() {
int result = 0; /* here *OUTER* result gets the value 0 */
while(result <= 0) /* THIS MAKES THE While to execute forever */
{
int result = (getchar() != EOF); /* THIS VARIABLE IS ***NOT*** THE outside result variable */
result = 2; /* external block result is not visible here so this assign goes to the above inner result */
printf("x");
/* INNER result CEASES TO EXIST HERE */
}
printf("outn");
}
As you can deduct from the comments, the result variable that is compared in the while test is the outer one, while the inner one hides the outer one, no assignations can be made to it in the body of the loop, so the loop runs forever. You get an infinite string of xs printed on stdout.
answered Nov 15 '18 at 9:41
Luis ColoradoLuis Colorado
4,3601718
add a comment |
Well,
#include <stdio.h>
int main() {
int result = 0; /* here *OUTER* result gets the value 0 */
while(result <= 0) /* THIS MAKES THE While to execute forever */
{
int result = (getchar() != EOF); /* THIS VARIABLE IS ***NOT*** THE outside result variable */
result = 2; /* external block result is not visible here so this assign goes to the above inner result */
printf("x");
/* INNER result CEASES TO EXIST HERE */
}
printf("outn");
}
As you can deduct from the comments, the result variable that is compared in the while test is the outer one, while the inner one hides the outer one, no assignations can be made to it in the body of the loop, so the loop runs forever. You get an infinite string of xs printed on stdout.
answered Nov 15 '18 at 9:41
Luis ColoradoLuis Colorado
4,3601718
add a comment |
Well,
#include <stdio.h>
int main() {
int result = 0; /* here *OUTER* result gets the value 0 */
while(result <= 0) /* THIS MAKES THE While to execute forever */
{
int result = (getchar() != EOF); /* THIS VARIABLE IS ***NOT*** THE outside result variable */
result = 2; /* external block result is not visible here so this assign goes to the above inner result */
printf("x");
/* INNER result CEASES TO EXIST HERE */
}
printf("outn");
}
As you can deduct from the comments, the result variable that is compared in the while test is the outer one, while the inner one hides the outer one, no assignations can be made to it in the body of the loop, so the loop runs forever. You get an infinite string of xs printed on stdout.
answered Nov 15 '18 at 9:41
Luis ColoradoLuis Colorado
4,3601718
Well,
#include <stdio.h>
int main() {
int result = 0; /* here *OUTER* result gets the value 0 */
while(result <= 0) /* THIS MAKES THE While to execute forever */
{
int result = (getchar() != EOF); /* THIS VARIABLE IS ***NOT*** THE outside result variable */
result = 2; /* external block result is not visible here so this assign goes to the above inner result */
printf("x");
/* INNER result CEASES TO EXIST HERE */
}
printf("outn");
}
As you can deduct from the comments, the result variable that is compared in the while test is the outer one, while the inner one hides the outer one, no assignations can be made to it in the body of the loop, so the loop runs forever. You get an infinite string of xs printed on stdout.
answered Nov 15 '18 at 9:41
Luis ColoradoLuis Colorado
4,3601718
answered Nov 15 '18 at 9:41
Luis ColoradoLuis Colorado
4,3601718
answered Nov 15 '18 at 9:41
Luis ColoradoLuis Colorado
4,3601718
answered Nov 15 '18 at 9:41
Luis ColoradoLuis Colorado
4,3601718
4,3601718
add a comment |
add a comment |
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1
Tip: a well enabled compiler may report "warning: variable 'result' set but not used [-Wunused-but-set-variable]". Save time and enable all compiler warnings.
– chux
Nov 9 '18 at 16:12