C++ 实例初始化语法 [英] C++ Instance Initialization Syntax

问题描述

给定这样的课程:

class Foo {
public:
    Foo(int);

    Foo(const Foo&);

    Foo& operator=(int);

private:
    // ...
};

这两行是完全等价的，还是它们之间有细微的差别?

Are these two lines exactly equivalent, or is there a subtle difference between them?

Foo f(42);

Foo f = 42;

<小时>

我通过在原始问题中使 Foo 构造函数显式"来混淆问题.我已将其删除，但感谢您的回答.

---

I confused matters by making the Foo constructor "explicit" in the original question. I've removed that, but appreciate the answers.

我还添加了复制构造函数的声明，以明确复制可能不是一个简单的操作.

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，按照C++标准，Foo f = 42"会直接调用Foo(int)构造函数，还是会调用拷贝构造函数?

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 fasih.ahmed 有我正在寻找的答案(除非它是错误的).

It looks like fasih.ahmed has the answer I was looking for (unless it'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);

该语句将直接赋值，因此少了一个函数调用.

This statement will directly assign the value so one less function call.

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