问题描述
给定类:
class Foo {
public:
Foo(int);
Foo(const Foo&);
Foo& operator=(int);
private:
// ...
};
这两行是完全相同的,还是它们之间有微妙的区别?
Are these two lines exactly equivalent, or is there a subtle difference between them?
Foo f(42);
Foo f = 42;
我通过使Foo构造函数在原始问题中明确而混淆了问题。我已经删除了,但感谢答案。
I confused matters by making the Foo constructor "explicit" in the original question. I've removed that, but appreciate the answers.
我还添加了一个复制构造函数的声明,以表明复制可能不是一个微不足道的操作。
I've also added declaration of a copy constructor, to make it clear that copying may not be a trivial operation.
我真正想知道的是,根据C ++标准,Foo f = 42直接调用Foo(int)构造函数,
What I really want to know is, according to the C++ standard, will "Foo f = 42" directly call the Foo(int) constructor, or is the copy constructor going to be called?
看起来像fasih.ahmed有我正在寻找的答案(除非他错了)。
It looks like fasih.ahmed has the answer I was looking for (unless he's wrong).
推荐答案
Foo f = 42;
此语句将为值'42'创建一个临时对象。
This statement will make a temporary object for the value '42'.
Foo f(42);
此语句将直接赋值,所以减少一个函数调用。
This statement will directly assign the value so one less function call.
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