为什么静态成员变量与三元运算符不匹配？ [英] Why don't static member variables play well with the ternary operator?

问题描述

这是协议。我有一个静态类，它包含用于获取输入的几个静态函数。该类包含一个私有静态成员变量，用于指示用户是否输入了任何信息。每个输入法检查用户是否输入任何信息，并相应地设置状态变量。我认为这将是一个好时机使用三元运算符。不幸的是，我不能，因为编译器不喜欢这样。

Here's the deal. I have a static class which contains several static functions used for getting input. The class contains a private static member variable for indicating whether the user entered any information. Each input method checks to see whether the user entered any information, and sets the status variable accordingly. I think this would be a good time to use the ternary operator. Unfortunately, I can't, because the compiler doesn't like that.

我复制了这个问题，然后简化我的代码尽可能多，理解。这不是我的原始代码。

I replicated the problem, then simplified my code as much as was possible to make it easy to understand. This is not my original code.

这是我的头文件：

#include <iostream>

using namespace std;

class Test {
public:
    void go ();
private:
    static const int GOOD = 0;
    static const int BAD = 1;
};

这是我的三元运算符的实现：

Here's my implementation with the ternary operator:

#include "test.h"

void Test::go () {
    int num = 3;
    int localStatus;
    localStatus = (num > 2) ? GOOD : BAD;
}

这里是主要功能：

#include <iostream>
#include "test.h"

using namespace std;

int main () {
    Test test = Test();
    test.go();
    return 0;
}

当我试图编译这个，我得到这个错误信息：

When I try to compile this, I get this error message:

test.o: In function `Test::go()':
test.cpp:(.text+0x17): undefined reference to `Test::GOOD'
test.cpp:(.text+0x1f): undefined reference to `Test::BAD'
collect2: ld returned 1 exit status

但是，如果我替换为：

localStatus = (num > 2) ? GOOD : BAD;

：

if (num > 2) {
    localStatus = GOOD;
} else {
    localStatus = BAD;
}

代码按预期编译和运行。什么模糊的C ++规则或GCC角的情况负责这种疯狂？ （我在Ubuntu 9.10上使用GCC 4.4.1。）

The code compiles and runs as expected. What obscure C++ rule or GCC corner case is responsible for this lunacy? (I'm using GCC 4.4.1 on Ubuntu 9.10.)

推荐答案

这是根据C ++标准。三元运算符构成一个单值，在运行时引用 GOOD 或 BAD 。左值到右值转换不会立即应用于左值 GOOD 或 BAD ，因此您需要定义 GOOD 和 BAD 。

This is according to the C++ Standard. The ternary operator does constitute a single lvalue that will refer to either GOOD or BAD at runtime. The lvalue to rvalue conversion is not applied immediately to either the lvalue GOOD or BAD, and therefor you require a definition of GOOD and BAD.

请参阅核心语言问题报告 http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_active.html#712 。

See the core language issue report http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_active.html#712 .

作为解决方法，您可以对 int （读取其值，从而对右值转换执行左值）应用显式转换或使用读取值的运算符，例如 + ：

As a workaround, you can apply explicit casts to int (which reads their values, thereby doing an lvalue to rvalue conversion) or use an operator that reads the value, like +:

localStatus = (num > 2) ? +GOOD : +BAD;

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