为什么这调用默认构造函数？ [英] Why does this call the default constructor?

问题描述

struct X
{
    X()    { std::cout << "X()\n";    }
    X(int) { std::cout << "X(int)\n"; }
};

const int answer = 42;

int main()
{
    X(answer);
}

我希望这可以打印

I would have expected this to print either

• X（int），因为 X（answer）; 可以解释为从 int 到 X 或

<没有什么，因为 X（answer）; 可以解释为一个变量的声明。

• X(int), because X(answer); could be interpreted as a cast from int to X, or
• nothing at all, because X(answer); could be interpreted as the declaration of a variable.

但是，它打印 X（） ，我有不知道为什么 X（answer）; 会调用默认构造函数。

However, it prints X(), and I have no idea why X(answer); would call the default constructor.

更改为获取临时而不是变量声明？

BONUS POINTS: What would I have to change to get a temporary instead of a variable declaration?

推荐答案

回答）;可以解释为一个变量的声明。

nothing at all, because X(answer); could be interpreted as the declaration of a variable.

你的答案隐藏在这里。如果你声明一个变量，你调用它的默认ctor（如果非POD和所有东西）。

Your answer is hidden in here. If you declare a variable, you invoke its default ctor (if non-POD and all that stuff).

在你的编辑：要获得一个临时，选项：

On your edit: To get a temporary, you have a few options:

• （answer））;


• （X）answer;


• static_cast< X>（answer）


• X {answer}; （C ++ 11）


• [] {return X（answer）; }（）; （C ++ 11，可能会产生副本）


• void（），X（answer）;


• X（（void（），answer））;


• X（answer）：X（）;


• if（X（answer），false）{}


• for（; X（answer），false;）;


• X（+ answer）;

• (X(answer));
• (X)answer;
• static_cast<X>(answer)
• X{answer}; (C++11)
• []{ return X(answer); }(); (C++11, may incur copy)
• void(), X(answer);
• X((void(),answer));
• true ? X(answer) : X();
• if(X(answer), false){}
• for(;X(answer), false;);
• X(+answer);

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