0
#include <stdio.h>
int main()
{
    int i=0;
    while(i++,i<=8);
    printf("%d\n",i);
    return 0;
}

Why is the increment of i done after the comparison in each test case?

Jonathan Leffler
  • 730,956
  • 141
  • 904
  • 1,278
Aarvi
  • 3
  • 3
  • is the `;` in `while` intentional? – Sourav Ghosh May 29 '18 at 13:00
  • `i` increases until the comparison is `false`. What is the first integer value for which that comparison would be `false`? – David May 29 '18 at 13:03
  • 5
    Why would you write the code this way? What's wrong with `while (i <= 8) i ++;`? – axiac May 29 '18 at 13:04
  • @axiax this is probably some sample code to demonstrate postincrement operator and maybe operator precedence of coma operator etc.. – Gerhardh May 29 '18 at 13:17
  • Note that the comma operator forces the increment to be complete _before_ (not after) each comparison. – Jonathan Leffler May 29 '18 at 13:24
  • "Why is the increment of i done after the comparison in each test case?" It isn't, and so you get the result 9? – Lundin May 29 '18 at 13:31
  • Note that `i++` means that i is incremented but the original value is returned. No matter how you write it the increment would always happen. And in your case the `,` is a sequence point so the increment and test are separated anyway. – Goswin von Brederlow May 29 '18 at 13:32

5 Answers5

8

i <= 8 succeeds for the last time when i = 8.

In the last iteration, i++ is executed, then i <= 8 fails because i = 9.

Note that the , is a sequence point, so i++ is absolutely guaranteed to be executed before i <= 8. Not so for similar constructs.

cadaniluk
  • 15,027
  • 2
  • 39
  • 67
2

It's not. Your loop condition is i <= 8, it is first non-true when i reaches 9 (you're incrementing i by 1 each time, so it will be exactly 9). That is why it prints 9.

bipll
  • 11,747
  • 1
  • 18
  • 32
2

To reach to the print() statement, while loop must end. The terminating condition, controlling expression should evaluate to false (or, in other words, until the controlling expression compares equal to 0), i.e., it will be false only when i <= 8; evaluates to false. For a value of i as 9, that happens.

Next line, the value of i, gets printed. So, you see 9.

Sourav Ghosh
  • 133,132
  • 16
  • 183
  • 261
1

Increment of i is not done after the comparison in each test case. i++ is executed first and after that the comparison is done.

This is because when expressions are separated using commas in C, the expressions are evaluated from left to right and the value of the last expression becomes the value of the total comma separated expression.

So the value of

i++,i<=8

is actually the value of i<=8 and that comparison is done only after i++ is executed.

So the while loop here

int i=0;
while(i++,i<=8);

is equivalent to

for(i=1; i<=8; i++);

Hence the control exits the loop only when i is greater than 8. Since i is incremented by 1 on each iteration, this means that the loop is over when i becomes 9.

J...S
  • 5,079
  • 1
  • 20
  • 35
0

You started i = 0 and using while ( i++, i<=8 ) loop you incremented it's value until 8 and when it's increased one more time i = 9 , then loop condition became false and breaks the loop with i = 9. And that's why now, when you print i's value, it gave you 9.

Innat
  • 16,113
  • 6
  • 53
  • 101