范围分辨率运算符 [英] Scope resolution operator

问题描述

我偶然发现这是在我正在看的源代码之一。

I accidentally happened to find this in one of the source codes I was looking at. So, I'm giving a similar smaller example here.

在 test.h b $ b

#include<iostream>

class test{
    int i;
public:
    test(){}
    //More functions here
};

在文件 test.cpp 中：

#include "test.h"

int main()
{
    test test1;
    test::test test2;
    test::test::test test3;
    return 0;
}

首先，是否有理由声明 test2 那样吗？其次，这个代码在g ++版本4.4.3和更低版本中编译得很好。在C ++标准中有一些东西，说，当不需要解析范围时，范围解析操作符被忽略。

First of all, is there a reason to declare test2 that way? Secondly, this code compiles just fine in g++ version 4.4.3 and lower versions. Is there something in the C++ standard, saying, scope resolution operators are ignored when there is no need to resolve scope?

推荐答案

代码无效。

这是g ++中的一个错误，它接受了代码。请参见g ++不正确处理注入的类名称。错误是解决为在2009年固定，因此它应该固定在任何最新版本的g ++。

It was a bug in g++ that it accepted the code. See "g++ does not treat injected class name correctly." The bug was resolved as fixed in 2009, so it should be fixed in any recent version of g++.

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