为什么以及如何GCC编译缺少return语句的函数吗？ [英] Why and how does GCC compile a function with a missing return statement?

问题描述

#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;
}

在TOUPPER函数，返回类型为char，但在没有TOUPPER回归（）。和编译源$ C ​​$ C与海湾合作委员会（GCC）4.5.1 20100924（红帽4.5.1-4）的Fedora-14。

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.

当然，发出警告：警告：控制​​到达非void函数结束，但是，做工精良

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

什么在code发生编译用gcc时？
我想在这种情况下，一个坚实的答案。
感谢：）

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

推荐答案

发生了什么事对你来说，当C程序被编译成汇编语言，你的toupper函数会弄成这个样子，也许是：

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

32的减法是在％eax寄存器进行。而在x86调用约定，就是在返回值预计将注册！所以...你真的很幸运。

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.

但请注意警告。他们是有原因的！

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