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, orX(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?
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);这篇关于为什么这会调用默认构造函数?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！