在复制构造函数中调用赋值运算符 [英] Calling assignment operator in copy constructor

问题描述

这种复制构造函数的实现有一些缺点吗？

Are there some drawbacks of such implementation of copy-constructor?

Foo::Foo(const Foo& i_foo)
{
   *this = i_foo;
}

正如我记得，在一些书中建议从赋值操作符和使用知名的交换技巧，但我不记得为什么...

As I remember, it was recommend in some book to call copy constructor from assignment operator and use well-known swap trick, but I don't remember, why...

推荐答案

是的，理念。用户定义类型的所有成员变量将首先被初始化，然后立即被覆盖。

Yes, that's a bad idea. All member variables of user-defined types will be initialized first, and then immediately overwritten.

这种交换技巧是：

Foo& operator=(Foo rhs) // note the copying
{
   rhs.swap(*this); //swap our internals with the copy of rhs
   return *this;
} // rhs, now containing our old internals, will be deleted 

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