本文介绍了为什么这会调用默认构造函数?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着跟版网的小编来一起学习吧！
问题描述
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), becauseX(answer);could be interpreted as a cast frominttoX, 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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