Question

I’m writting a small code, just five lines, but it can’t runs as the correct order.
After compiling this code, It will exec "dir" first and then print my string.

But if I change it to printf ("%s\n", "asdf");, the program will run with correct order.
And I want to know why it will be this. (PS, my computer is arch with gcc 4.6.2 and I also use clang but they have the same result.)
Thank you every one.
Here is the code:

#include <stdio.h>
#include <stdlib.h>
int main()
{
    printf ("%s", "asdf");
    system ("dir");

    return 0;
} 

Answer 1

The reason it looks like dir is being executed before the printf is because printf prints to stdout and stdout buffers its output – your program is being executed in the correct order. stdout‘s buffer is usually flushed when (as I mentioned in this answer):

• A newline ('\n') is to be printed
• You read in from stdin
• fflush() is called on it

That is why adding a newline ('\n') the string being printed causes "asdf" to be printed before the output of dir.

Other alternatives to flushing the buffer include printing to stderr, which isn’t buffered, or using setbuf(stdout, NULL); to disable buffering altogether for stdout. Neither of these two approaches seem necessary for your situation, however.