How while loop works with two conditions in c ??

Question

while(a<7,b<17)
{
a++;
b++;
}
printf("%d %d",a,b);
the answer of this challenge is 17,17 don't know how 🤔🤔

Answer 1

I'd either go for a<7&&b<17 or a<7||b<17. &&: both conditions must be true, ||: it stops as soon as one of the conditions is true

Answer 2

The comma operator evaluates the first expression, throws the value away, and then evaluates the second expression.

if (x = 0)

will not execute, but

if (x = 0, x >-1) will.

You only use , when the first expression *does* something (has a side effect).

Answer 3

very thanks

Answer 4

I think I understand the question now.

If the answer is 17, 17
the loop should be

while((a<7) || (b<17)){
a++;
b++;
}

This means the loop runs as long as either of the conditions are true. Since both are less than statements only the higher number would matter.

Same thing could be accomplished with

while(b<17){
a++;
b++;
}

If && was used instead of || then the output would be 7 7.

Answer 5

You could add two if statements to stop a at 7 and b at 17

int main(void){
int a, b = 0;

while((a<7) || (b<17){
if(a < 7){
a++;
}
if(b < 17){
b++;
}
return 0;
}
}

Answer 6

While(a<7 || b<17)
//that for "or"

While(a<7 && b<17)
//that for "and"