For Loop with Multivariable in Kotlin
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How can I do this java code in kotlin using just one for loop?
for(int i=0, j=0; i < 6 && j < 6; i++, j+=2) {
// code here
}
kotlin
asked Nov 16 '18 at 14:39
mosyonalmosyonal
134
add a comment |
How can I do this java code in kotlin using just one for loop?
for(int i=0, j=0; i < 6 && j < 6; i++, j+=2) {
// code here
}
kotlin
asked Nov 16 '18 at 14:39
mosyonalmosyonal
134
you need just one variable for such a loop
– Maxim
Nov 16 '18 at 18:35
add a comment |
How can I do this java code in kotlin using just one for loop?
for(int i=0, j=0; i < 6 && j < 6; i++, j+=2) {
// code here
}
kotlin
asked Nov 16 '18 at 14:39
mosyonalmosyonal
134
How can I do this java code in kotlin using just one for loop?
for(int i=0, j=0; i < 6 && j < 6; i++, j+=2) {
// code here
}
kotlin
kotlin
asked Nov 16 '18 at 14:39
mosyonalmosyonal
134
asked Nov 16 '18 at 14:39
mosyonalmosyonal
134
asked Nov 16 '18 at 14:39
mosyonalmosyonal
134
asked Nov 16 '18 at 14:39
mosyonalmosyonal
134
asked Nov 16 '18 at 14:39
mosyonalmosyonal
134
134
you need just one variable for such a loop
– Maxim
Nov 16 '18 at 18:35
add a comment |
you need just one variable for such a loop
– Maxim
Nov 16 '18 at 18:35
you need just one variable for such a loop
– Maxim
Nov 16 '18 at 18:35
you need just one variable for such a loop
– Maxim
Nov 16 '18 at 18:35
add a comment |
2 Answers
2
active
oldest
votes
There is no way to iterate over multiple variables. In this case, the easiest thing you can do is:
for (i in 0..3) {
val j = i * 2
}
In a more general case, you can rewrite this as a while loop:
var i = 0
var j = 0
while (i < 6 && j < 6) {
// code here
i++
j += 2
}
answered Nov 16 '18 at 14:42
yoleyole
61.7k11159149
yeah I know these, I think maybe there is a way but I can't find it so I asked here. Thank you!
– mosyonal
Nov 16 '18 at 15:16
The Java to Kotlin conversion tool converts it to a while loop too.
– Zoe
Nov 16 '18 at 17:15
i < 6is redundant here. actually it's clear from the first code block
– Maxim
Nov 16 '18 at 18:38
@Maxim I know, thanks. I'm showing how to rewrite the code in the general case when there is no such dependency between the two loop variables.
– yole
Nov 17 '18 at 8:59
@yole ah now I see
– Maxim
Nov 17 '18 at 19:03
add a comment |
yole's answer is almost certainly the simplest and most efficient approach.
But one alternative you might look at is zipping sequences together, e.g.:
for ((i, j) in sequence{ yieldAll(0 until 6) }.zip(sequence{ yieldAll(0 until 6 step 2) })) {
// code here
}
That's much more readable with a utility function, e.g.:
fun <T, U> seqs(it1: Iterable<T>, it2: Iterable<U>)
= sequence{ yieldAll(it1) }.zip(sequence{ yieldAll(it2) })
for ((i, j) in seqs(0 until 6, 0 until 6 step 2)) {
// code here
}
That's less efficient (initially creating Iterables, Ranges, and Sequences, and then a Pair for each iteration). But it's the exact equivalent of the code in the question. And because it defines each range in one place, it does at least make them very clear.
(I think this needs Kotlin 1.3. There are probably simpler and/or more general ways of doing it; feel free to edit this if you can improve it!)
answered Nov 16 '18 at 17:17
giddsgidds
1,483128
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
There is no way to iterate over multiple variables. In this case, the easiest thing you can do is:
for (i in 0..3) {
val j = i * 2
}
In a more general case, you can rewrite this as a while loop:
var i = 0
var j = 0
while (i < 6 && j < 6) {
// code here
i++
j += 2
}
answered Nov 16 '18 at 14:42
yoleyole
61.7k11159149
yeah I know these, I think maybe there is a way but I can't find it so I asked here. Thank you!
– mosyonal
Nov 16 '18 at 15:16
The Java to Kotlin conversion tool converts it to a while loop too.
– Zoe
Nov 16 '18 at 17:15
i < 6is redundant here. actually it's clear from the first code block
– Maxim
Nov 16 '18 at 18:38
@Maxim I know, thanks. I'm showing how to rewrite the code in the general case when there is no such dependency between the two loop variables.
– yole
Nov 17 '18 at 8:59
@yole ah now I see
– Maxim
Nov 17 '18 at 19:03
add a comment |
There is no way to iterate over multiple variables. In this case, the easiest thing you can do is:
for (i in 0..3) {
val j = i * 2
}
In a more general case, you can rewrite this as a while loop:
var i = 0
var j = 0
while (i < 6 && j < 6) {
// code here
i++
j += 2
}
answered Nov 16 '18 at 14:42
yoleyole
61.7k11159149
yeah I know these, I think maybe there is a way but I can't find it so I asked here. Thank you!
– mosyonal
Nov 16 '18 at 15:16
The Java to Kotlin conversion tool converts it to a while loop too.
– Zoe
Nov 16 '18 at 17:15
i < 6is redundant here. actually it's clear from the first code block
– Maxim
Nov 16 '18 at 18:38
@Maxim I know, thanks. I'm showing how to rewrite the code in the general case when there is no such dependency between the two loop variables.
– yole
Nov 17 '18 at 8:59
@yole ah now I see
– Maxim
Nov 17 '18 at 19:03
add a comment |
There is no way to iterate over multiple variables. In this case, the easiest thing you can do is:
for (i in 0..3) {
val j = i * 2
}
In a more general case, you can rewrite this as a while loop:
var i = 0
var j = 0
while (i < 6 && j < 6) {
// code here
i++
j += 2
}
answered Nov 16 '18 at 14:42
yoleyole
61.7k11159149
There is no way to iterate over multiple variables. In this case, the easiest thing you can do is:
for (i in 0..3) {
val j = i * 2
}
In a more general case, you can rewrite this as a while loop:
var i = 0
var j = 0
while (i < 6 && j < 6) {
// code here
i++
j += 2
}
answered Nov 16 '18 at 14:42
yoleyole
61.7k11159149
answered Nov 16 '18 at 14:42
yoleyole
61.7k11159149
answered Nov 16 '18 at 14:42
yoleyole
61.7k11159149
answered Nov 16 '18 at 14:42
yoleyole
61.7k11159149
61.7k11159149
yeah I know these, I think maybe there is a way but I can't find it so I asked here. Thank you!
– mosyonal
Nov 16 '18 at 15:16
The Java to Kotlin conversion tool converts it to a while loop too.
– Zoe
Nov 16 '18 at 17:15
i < 6is redundant here. actually it's clear from the first code block
– Maxim
Nov 16 '18 at 18:38
@Maxim I know, thanks. I'm showing how to rewrite the code in the general case when there is no such dependency between the two loop variables.
– yole
Nov 17 '18 at 8:59
@yole ah now I see
– Maxim
Nov 17 '18 at 19:03
add a comment |
yeah I know these, I think maybe there is a way but I can't find it so I asked here. Thank you!
– mosyonal
Nov 16 '18 at 15:16
The Java to Kotlin conversion tool converts it to a while loop too.
– Zoe
Nov 16 '18 at 17:15
i < 6is redundant here. actually it's clear from the first code block
– Maxim
Nov 16 '18 at 18:38
@Maxim I know, thanks. I'm showing how to rewrite the code in the general case when there is no such dependency between the two loop variables.
– yole
Nov 17 '18 at 8:59
@yole ah now I see
– Maxim
Nov 17 '18 at 19:03
yeah I know these, I think maybe there is a way but I can't find it so I asked here. Thank you!
– mosyonal
Nov 16 '18 at 15:16
yeah I know these, I think maybe there is a way but I can't find it so I asked here. Thank you!
– mosyonal
Nov 16 '18 at 15:16
The Java to Kotlin conversion tool converts it to a while loop too.
– Zoe
Nov 16 '18 at 17:15
The Java to Kotlin conversion tool converts it to a while loop too.
– Zoe
Nov 16 '18 at 17:15
i < 6 is redundant here. actually it's clear from the first code block– Maxim
Nov 16 '18 at 18:38
i < 6 is redundant here. actually it's clear from the first code block– Maxim
Nov 16 '18 at 18:38
@Maxim I know, thanks. I'm showing how to rewrite the code in the general case when there is no such dependency between the two loop variables.
– yole
Nov 17 '18 at 8:59
@Maxim I know, thanks. I'm showing how to rewrite the code in the general case when there is no such dependency between the two loop variables.
– yole
Nov 17 '18 at 8:59
@yole ah now I see
– Maxim
Nov 17 '18 at 19:03
@yole ah now I see
– Maxim
Nov 17 '18 at 19:03
add a comment |
yole's answer is almost certainly the simplest and most efficient approach.
But one alternative you might look at is zipping sequences together, e.g.:
for ((i, j) in sequence{ yieldAll(0 until 6) }.zip(sequence{ yieldAll(0 until 6 step 2) })) {
// code here
}
That's much more readable with a utility function, e.g.:
fun <T, U> seqs(it1: Iterable<T>, it2: Iterable<U>)
= sequence{ yieldAll(it1) }.zip(sequence{ yieldAll(it2) })
for ((i, j) in seqs(0 until 6, 0 until 6 step 2)) {
// code here
}
That's less efficient (initially creating Iterables, Ranges, and Sequences, and then a Pair for each iteration). But it's the exact equivalent of the code in the question. And because it defines each range in one place, it does at least make them very clear.
(I think this needs Kotlin 1.3. There are probably simpler and/or more general ways of doing it; feel free to edit this if you can improve it!)
answered Nov 16 '18 at 17:17
giddsgidds
1,483128
add a comment |
yole's answer is almost certainly the simplest and most efficient approach.
But one alternative you might look at is zipping sequences together, e.g.:
for ((i, j) in sequence{ yieldAll(0 until 6) }.zip(sequence{ yieldAll(0 until 6 step 2) })) {
// code here
}
That's much more readable with a utility function, e.g.:
fun <T, U> seqs(it1: Iterable<T>, it2: Iterable<U>)
= sequence{ yieldAll(it1) }.zip(sequence{ yieldAll(it2) })
for ((i, j) in seqs(0 until 6, 0 until 6 step 2)) {
// code here
}
That's less efficient (initially creating Iterables, Ranges, and Sequences, and then a Pair for each iteration). But it's the exact equivalent of the code in the question. And because it defines each range in one place, it does at least make them very clear.
(I think this needs Kotlin 1.3. There are probably simpler and/or more general ways of doing it; feel free to edit this if you can improve it!)
answered Nov 16 '18 at 17:17
giddsgidds
1,483128
add a comment |
yole's answer is almost certainly the simplest and most efficient approach.
But one alternative you might look at is zipping sequences together, e.g.:
for ((i, j) in sequence{ yieldAll(0 until 6) }.zip(sequence{ yieldAll(0 until 6 step 2) })) {
// code here
}
That's much more readable with a utility function, e.g.:
fun <T, U> seqs(it1: Iterable<T>, it2: Iterable<U>)
= sequence{ yieldAll(it1) }.zip(sequence{ yieldAll(it2) })
for ((i, j) in seqs(0 until 6, 0 until 6 step 2)) {
// code here
}
That's less efficient (initially creating Iterables, Ranges, and Sequences, and then a Pair for each iteration). But it's the exact equivalent of the code in the question. And because it defines each range in one place, it does at least make them very clear.
(I think this needs Kotlin 1.3. There are probably simpler and/or more general ways of doing it; feel free to edit this if you can improve it!)
answered Nov 16 '18 at 17:17
giddsgidds
1,483128
yole's answer is almost certainly the simplest and most efficient approach.
But one alternative you might look at is zipping sequences together, e.g.:
for ((i, j) in sequence{ yieldAll(0 until 6) }.zip(sequence{ yieldAll(0 until 6 step 2) })) {
// code here
}
That's much more readable with a utility function, e.g.:
fun <T, U> seqs(it1: Iterable<T>, it2: Iterable<U>)
= sequence{ yieldAll(it1) }.zip(sequence{ yieldAll(it2) })
for ((i, j) in seqs(0 until 6, 0 until 6 step 2)) {
// code here
}
That's less efficient (initially creating Iterables, Ranges, and Sequences, and then a Pair for each iteration). But it's the exact equivalent of the code in the question. And because it defines each range in one place, it does at least make them very clear.
(I think this needs Kotlin 1.3. There are probably simpler and/or more general ways of doing it; feel free to edit this if you can improve it!)
answered Nov 16 '18 at 17:17
giddsgidds
1,483128
answered Nov 16 '18 at 17:17
giddsgidds
1,483128
answered Nov 16 '18 at 17:17
giddsgidds
1,483128
answered Nov 16 '18 at 17:17
giddsgidds
1,483128
1,483128
add a comment |
add a comment |
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you need just one variable for such a loop
– Maxim
Nov 16 '18 at 18:35