为什么<而不是<在流输出仍然编译？ [英] Why does < instead of << in stream output still compile?

问题描述

今天我在我的程序中做了一个小的打字错误，并且正在徘徊为什么我没有得到任何输出，虽然程序编译良好。基本上它减少到这：

Today I made a small typo in my program, and was wandering why I wasn't getting any output, although the program compiled fine. Basically it reduces to this:

#include <iostream>

int main()
{
    std::cout < "test"; // no << but <
}

我完全不知道这里执行什么样的隐式转换，所以程序仍然编译（g ++ 4.9.2和甚至g ++ 5）。我只是意识到clang ++拒绝代码。是否有转换到 void * 执行（不能想到别的）？我记得看到这样的东西，但我认为这是在g ++ 5中解决，但似乎不是这样的情况。

I have absolutely no idea what kind of implicit conversion is performed here so the program still compiles (both g++4.9.2 and even g++5). I just realized that clang++ rejects the code. Is there a conversion to void* being performed (cannot think of anything else)? I remember seeing something like this, but I thought it was addressed in g++5, but this doesn't seem to be the case.

编辑：我没有使用 -std = c ++ 11 所以代码在前C ++ 11中是有效的（由于转换为 void * of ostream ）。当用 -std = c ++ 11 g ++ 5拒绝代码编译时，g ++ 4.9仍然接受它。

I was not compiling with -std=c++11, so the code was valid in pre-C++11 (due to conversion to void* of ostream). When compiling with -std=c++11 g++5 rejects the code, g++4.9 still accepts it.

推荐答案

是的，编译器将 cout 转换为 void * 。如果你使用 -S 开关来获取代码的反汇编，你会看到这样的：

Yes, the compiler is converting cout to a void*. If you use the -S switch to get the code's disassembly, you'll see something like this:

    mov edi, OFFSET FLAT:std::cout+8
    call    std::basic_ios<char, std::char_traits<char> >::operator void*() const
    cmp rax, OFFSET FLAT:.LC0
    setb    al
    test    al, al

这说明 operator void * 是罪魁祸首。

与Bill Lynch所说的相反，我可以在 -std = c ++ 11 goo.gl/RgweWSrel =nofollow>编译器资源管理器。但是，它似乎是一个实现缺陷，因为C ++ 11应该替换 operator void * 和 operator bool basic_ios 。

Contrary to what Bill Lynch said, I'm able to reproduce it with —std=c++11 on Compiler Explorer. However, it does appear to be an implementation defect, since C++11 should have replaced operator void* with operator bool on basic_ios.

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