Some one can explain this behavior ? just 2 lines of code [duplicate]

Question

Kindly explain this snippet:

#include <stdio.h>

int puts(const char *str) {
    fputs("Hello world!\n", stdout);
}

int main() {
    printf("Goodbye\n");
}

Output : Hello world! return 13

Answer 1

It is compiler specific. You get this behavior with GCC. Here are some details.

• since you #include <stdio.h> (actually because you are in a hosted environment) the puts is that of the C99 standard, and redefining it is undefined behavior

• the GCC compiler has some optimizations to transform some printf to a sequence of faster puts. This is legal, since you have included <stdio.h> (and the C99 standard defines what printf should do in that case; GCC goes thru __builtin_printf as an intermediate step)

If you compile with -ffreestanding you won't observe that.

Your question is very close to this one; so this answer is also relevant.

Answer 2

I compiled the program with gcc x.c -S -o-. It gave me

[...]
main:
.LFB1:
        .cfi_startproc
        pushl   %ebp
        .cfi_def_cfa_offset 8
        .cfi_offset 5, -8
        movl    %esp, %ebp
        .cfi_def_cfa_register 5
        andl    $-16, %esp
        subl    $16, %esp
        movl    $.LC1, (%esp)
        call    puts
        leave
        .cfi_restore 5
        .cfi_def_cfa 4, 4
        ret
        .cfi_endproc
.LFE1:

so indeed the printf call is replaced with puts in GCC, as they have the same semantics.

Answer 3

My guess would be that the compiler changes the call to printf() into a call to puts(), since there's no need for printf() due to no formatting. Also the string is terminated by a newline, which fits puts(). The compiler didn't see your scary overload of a library function coming, so it got "fooled".