为什么这会调用默认构造函数？

Question

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

const int answer = 42;

int main()
{
    X(answer);
}

I would have expected this to print either

• 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.

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

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

Answer 1

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

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

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

• (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);

Answer 2

The parentheses are optional. What you said is identical to X answer; , and it's a declaration statement.

Answer 3

如果要声明X类型的变量，则应该这样做：

X y(answer);