Why and how does GCC compile a function with a missing return statement?

Question

#include <stdio.h>

char toUpper(char);

int main(void)
{
    char ch, ch2;
    printf("lowercase input : ");
    ch = getchar();
    ch2 = toUpper(ch);
    printf("%c ==> %c\n", ch, ch2);

    return 0;
}

char toUpper(char c)
{
    if(c>='a'&&c<='z')
        c = c - 32;
}

In toUpper function, return type is char, but there is no "return" in toUpper(). And compile the source code with gcc (GCC) 4.5.1 20100924 (Red Hat 4.5.1-4), fedora-14.

Of course, warning is issued: "warning: control reaches end of non-void function", but, working well.

What has happened in that code during compile with gcc? I want to get a solid answer in this case. Thanks :)

Answer 1

You should keep in mind that such code may crash depending on compiler. For example, clang generates ud2 instruction at the end of such function and your app will crash at run-time.

Answer 2

What happened for you is that when the C program was compiled into assembly language, your toUpper function ended up like this, perhaps:

_toUpper:
LFB4:
        pushq   %rbp
LCFI3:
        movq    %rsp, %rbp
LCFI4:
        movb    %dil, -4(%rbp)
        cmpb    $96, -4(%rbp)
        jle     L8
        cmpb    $122, -4(%rbp)
        jg      L8
        movzbl  -4(%rbp), %eax
        subl    $32, %eax
        movb    %al, -4(%rbp)
L8:
        leave
        ret

The subtraction of 32 was carried out in the %eax register. And in the x86 calling convention, that is the register in which the return value is expected to be! So... you got lucky.

But please pay attention to the warnings. They are there for a reason!