why doesn't the second while loop run?

Question

I have the following code, but why doesn't the second while loop run? Any changes to make it run?

#include<stdio.h>
int main() {
    int a,b,c,d,e,f;
    while(scanf("%d,%d,%d",&a,&b,&c)==3){
        printf("ok\n");
    }
    while(scanf("%d,%d,%d",&d,&e,&f)==3){
        printf("OK\n");
    }
    return 0;
}

My input is

1,2,3
1,5,7,4,8,7
7,8,9,...

1,2,3(press enter) 1,5,7, stops at this point but needs to take the input 4,8,7 (press enter) 7,8,9,.. — hrithik mahesh
– hrithik mahesh, Commented Mar 13, 2021 at 11:19
the second loop won't run until the first one is running. try to enter less numbers and see that the first loop will stop and the second will run. — MarcoLucidi
– MarcoLucidi, Commented Mar 13, 2021 at 11:22
when it enter a,b,c it keeps running but when I enter a,b,c, it will stop and second while should run but it's not running — hrithik mahesh
– hrithik mahesh, Commented Mar 13, 2021 at 11:24
Re "when I enter a,b,c, it will stop and second while should run but it's not running", That's because you have ,4,8,7 but your pattern expects 4,8,7 — ikegami
– ikegami, Commented Mar 13, 2021 at 11:28

pmg · Accepted Answer · 2021-03-13 11:37:09Z

1

Let's say you input "1,5,7,4,8,7<ENTER>" for the scanf() in the 1st while.

The scanf("%d,%d,%d", &a, &b, &c); reads 1 into a, 5 into b, 7 into c and returns 3 keeping ",4,8,7<ENTER>" in the input buffer.

In the 2nd time through the loop, scanf finds the comma and returns 0 which terminates the while.

Immediately after that, in the 2nd while, the 2nd scanf() attempts to convert ",4,8,7<ENTER>" to an integer and fails promptly returning 0 and terminating the while.

edited Mar 13, 2021 at 11:37

answered Mar 13, 2021 at 11:31

pmg

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