为什么在C ++中显式调用构造函数 [英] Why explicitly call a constructor in C++

问题描述

我知道我们可以使用范围解析操作符，即 className :: className（）显式调用C ++中的类的构造函数。

I know we can explicitly call the constructor of a class in C++ using scope resolution operator, i.e. className::className(). I was wondering where exactly would I need to make such a call.

推荐答案

最常见的是，在一个子类构造函数中需要一些参数：

Most often, in a child class constructor that require some parameters :

class BaseClass
{
public:
    BaseClass( const std::string& name ) : m_name( name ) { }

    const std::string& getName() const { return m_name; }

private:

    const std::string m_name;

//...

};


class DerivedClass : public BaseClass
{
public:

    DerivedClass( const std::string& name ) : BaseClass( name ) { }

// ...
};

class TestClass : 
{
public:
    TestClass( int testValue ); //...
};

class UniqueTestClass 
     : public BaseClass
     , public TestClass
{
public:
    UniqueTestClass() 
       : BaseClass( "UniqueTest" ) 
       , TestClass( 42 )
    { }

// ...
};

...例如。

除此之外，我没有看到效用。我只是在其他代码中调用构造函数，当我太年轻，不知道我在做什么...

Other than that, I don't see the utility. I only did call the constructor in other code when I was too young to know what I was really doing...

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