当函数中没有指定返回值时，C ++程序如何获得它们的返回值？ [英] How do C++ progs get their return value, when a return is not specified in the function?

问题描述

我最近写了一篇文章：

http://stackoverflow.com/questions/3452355/weird-error-in-c-program-removing-printout-breaks-program

...在其中我试图解决一个看似困惑的问题，其中删除cout语句会打破我的程序。

...in which I was trying to solve a seemingly baffling problem, in which removing a cout statement would break my program.

原来，我的问题是，我忘了返回我的真/假成功标志，我以后用于逻辑。

As it turned out, my problem was that I forgot to return my true/false success flag that I was later using for logic.

但显然SOMETHING被返回，

But apparently SOMETHING was being returned and that something was always true if I left that cout in, but would seemingly "magically" become false when I took it out.

我对你的问题是：
什么决定什么是c ++函数返回时没有返回命令在函数内执行？它有什么逻辑吗？

显然，忘记你的返回类型是一个坏主意。在这种情况下，虽然，这主要是由于我的程序的性质 - 一个快速的黑客工作。我后来决定，不值得努力包括实现一个算法来确定函数调用的成功/失败 - 但意外留下的代码依赖于返回。

Obviously forgetting your return type is a bad idea. In this case, though, it was largely due to the nature of my program -- a quick hack job. I later decided that it wasn't worth the effort to include implement an algorithm to determine the success/failure of the function call -- but accidentally left behind the code dependent on the return.

Bafflingly g ++在编译可执行文件时没有给出警告或错误：

Bafflingly g++ gave me no warnings or errors when compiling the executable like so:

g++ main.cc -g -o it_util

我的版本是：
g ++（GCC） 4.1.2 20080704 （Red Hat 4.1.2-44）

My version is: g++ (GCC) 4.1.2 20080704 (Red Hat 4.1.2-44)

再次，为了拯救别人，如果他们犯同样的愚蠢错误，不规则的行为，任何人都可以轻松地在一个没有返回的函数的返回值从

Again, to save others future frustration in case they make the same silly mistake and are met with the same seemingly erratic behavior, can anyone cast light on where a function without a return gets its return value from??

谢谢!!

推荐答案

在x86调用约定上，整数和指针的返回值在EAX寄存器中。下面是一个例子：

On x86 calling conventions, the return value for integers and pointers is on the EAX register. The following is an example of that:

int func() {
    if(0) return 5; // otherwise error C4716: 'func' : must return a value
}
int main() {
    int a;
    a = func();
}

使用 cl.exe / Zi ，MSVC ++ 10：

Compiling with cl.exe /Zi, MSVC++10:

push    ebp
mov     ebp, esp
push    ecx
call    j_?func@@YAHXZ  ; func(void)
mov     [ebp+a], eax ; assumes eax contains the return value
xor     eax, eax
mov     esp, ebp
pop     ebp
retn

当然，这是所有未定义的行为。

Of course, this is all undefined behavior.

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