quapka quapka - 2 years ago 225
C Question

C/C++ printf() before scanf() issue

I'm using Eclipse to code in C/C++ and I'm struggling with what might be something pretty easy. In my code below I use

printf()
and after
scanf()
. Althougth
printf
is written before
scanf()
the output differs. I was able to find out something about similar issue here. But I wasn't able to solve it. Any ideas?

Code:

#include <stdio.h>

int main()
{
int myvariable;

printf("Enter a number:");
scanf("%d", &myvariable);
printf("%d", myvariable);

return 0;
}


Expected output:

Enter a number:1
1


Instead I get:

1
Enter a number:1

Answer Source

Your output is being buffered. You have 4 options:

  1. explicit flush

    fflush after each write to profit from the buffer and still enforce the desiredbehavior/display explicitly.

    fflush( stdout );
    
  2. have the buffer only buffer lines-wise

    useful for when you know that it is enough to print only complete lines

    setlinebuf(stdout);
    
  3. disable the buffer

    setbuf(stdout, NULL);
    
  4. disable buffering in your console through what ever options menu it provides


Examples:

Here is your code with option 1:

#include <stdio.h>
int main() {

    int myvariable;

    printf("Enter a number:");
    fflush( stdout );
    scanf("%d", &myvariable);
    printf("%d", myvariable);
    fflush( stdout );

    return 0;
}

Here is 2:

#include <stdio.h>
int main() {

    int myvariable;

    setlinebuf(stdout);    

    printf("Enter a number:");
    scanf("%d", &myvariable);
    printf("%d", myvariable);

    return 0;
}

and 3:

#include <stdio.h>
int main() {

    int myvariable;

    setbuf(stdout, NULL);     

    printf("Enter a number:");
    scanf("%d", &myvariable);
    printf("%d", myvariable);

    return 0;
}
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