为什么这会调用默认构造函数?

2021-12-29 00:00:00 syntax c++ most-vexing-parse

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

const int answer = 42;

int main()
{
    X(answer);
}

我原以为这会打印

X(int)，因为 X(answer); 可以解释为从 int 到 X，或
什么都没有，因为 X(answer); 可以解释为变量的声明.



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.

然而，它打印X()，我不知道为什么X(answer); 将调用默认构造函数.

However, it prints X(), and I have no idea why X(answer); would call the default constructor.
奖励积分:我需要更改什么才能获得临时声明而不是变量声明?

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

什么都没有，因为 X(answer);可以解释为变量的声明.


  nothing at all, because X(answer); could be interpreted as the declaration of a variable.
您的答案隐藏在这里.如果你声明一个变量，你就会调用它的默认构造函数(如果是非 POD 和所有这些东西).

Your answer is hidden in here. If you declare a variable, you invoke its default ctor (if non-POD and all that stuff).
在您的编辑中:要获得临时文件，您有几个选择:

On your edit: To get a temporary, you have a few options:
(X(answer));
(X)answer;
static_cast(答案)
X{answer}; (C++11)
[]{ return X(answer);}();(C++11，可能会导致复制)
void(), X(answer);
X((void(),answer));
true ?X(answer) : X();
if(X(answer), false){}
for(;X(answer), false;);
X(+answer);


	
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