printf is not working as expected. Can someone explain the output?

Question

I am trying to read two strings from the keyboard and print them. Why does printf("read 1st\n") run after the second scanf()?

#include <stdio.h>

int main(void)
{
    char str[20];
    char str2[20];

    scanf("%s", str);
    printf("read 1st\n");
    scanf("%s", str2);
    printf("read 2nd\n");

    printf("str: %s\nstr2: %s\n", str, str2);

    return 0;
}

Expected:
foo
read 1st
bar
read 2nd
str: foo
str2: bar

Actual:
foo
bar
read 1st
read 2nd
str:foo
str2:bar

Can't reproduce. If I use your code I get the result you expect. Which compiler do you use? Which platform do you use? — user6556709, May 05 '19 at 11:10
Possible duplicate of [C/C++ printf() before scanf() issue](https://stackoverflow.com/questions/16877264/c-c-printf-before-scanf-issue) — Acorn, May 05 '19 at 11:11
I'm using windows 10 and the gcc compiler found here https://sourceforge.net/projects/tdm-gcc/ — Stefanos Poriazis, May 05 '19 at 11:14
It is more of a problem of *how you run* the program. Some IDEs are known to be broken in this regard. — Antti Haapala -- Слава Україні, May 05 '19 at 11:15
The default is that the code should behave as *you* expect unless the program thinks it is **not** connected to an interactive terminal. And this happens for example when there is a stupid IDE intervening with the interactive session. — Antti Haapala -- Слава Україні, May 05 '19 at 11:16

score 1 · Answer 1 · answered May 05 '19 at 11:10

I could not reproduce your problem, but adding a fflush(stdout); should take care of your problem.

scanf("%s", str);
printf("read 1st\n");
fflush(stdout); // Ensures that the above is printed before scanf is executed
scanf("%s", str2);
printf("read 2nd\n");

I would also change the scans to scanf("%19s", str);. Otherwise, bad things could happen if you enter a string longer than str can hold.

printf is not working as expected. Can someone explain the output?

1 Answers1