C line order reversed by compiler?

Question

I've got this sample code:

#include <stdio.h>
#include <stdlib.h>

int addOne();

int main(int argc, char** argv) {
    char s[100];
    int x = 1;
    x = addOne(x);
    printf("%d",x);
    gets(s);
    return (EXIT_SUCCESS);
}

int addOne(int j) {
    return j + 1;
}

You can see that printf is before gets(s), but when I run the code it asks the string first, after it prints the number. I'm using NetBeans with Cygwin gcc compiler. Is there something I miss? Is it a compiler error?

Answer 1

printf() is usually line-buffered. So printf's output doesn't appear on the screen until the buffer is flushed.

Add \n to the printf and it should work as expected.

printf("%d\n",x);

You can disable the buffering if it's not needed using setbuf(stdout, NULL);.

By the way, you should never gets(). Use fgets() instead. Because gets() is prone buffer overrun vulnerability and has been obsoleted since C99 and has been removed completely from C11.

---

Regarding your suspicion that compiler is re-ordering the lines: It's true that compiler can re-order instructions and is allowed to do so legally. But only as long as it doesn't affect observable effect of the program.

Consider the following code,

void func(void)
{
   int a = 2;
   int b = 3;

   a = a * 2;  //line 1
   b = b + 3;  //line 2

   printf("%d\n", a+b);
}

Compiler can compute either a or b in any order (line1 or line2) as it wouldn't affect the behaviour of the code. Or it can even simply replace the function with:

void func(void)
{
   printf("%d\n", 10);
}

Typically, compiler would do such transformations for optimization purpose and is allowed. In your code, compiler can't do such code transformations as it would affect the observable behaviour of your code.