显式复制构造函数 [英] Explicit copy constructor

问题描述

我已经扩展了std :: string，以满足我将自定义函数编译到 CustomString 字符串类中的需要

I have extended std::string to fulfil my needs of having to write custom function build into string class called CustomString

定义的构造函数：

I have defined constructors:

    class CustomString : public std::string {
    public:
        explicit CustomString(void);
        explicit CustomString(const std::string& str);
        explicit CustomString(const CustomString& customString);
        //assignment operator
        CustomString& operator=(const CustomString& customString);
    ... };

在第三个构造函数（复制构造函数）和赋值运算符中，它的定义是：

In the third constructor (copy constructor) and assignment operator, whose definition is:

CustomString::CustomString(const CustomString& customString):
    std::string(static_cast<std::string>(customString)) 
{}
CustomString& CustomString::operator=(const CustomString& customString){
    this->assign(static_cast<std::string>(customString));
    return *this;
}

首先，因为这是一个显式这意味着需要显式转换来分配给另一个CustomString对象;它抱怨作业。

First since this is an "explicit"; meaning an explicit cast is needed to assign to another CustomString object; it's complaining about the assignment.

CustomString s = CustomString("test");

我不知道明确地在哪里投放。

I am not sure where exactly is casting needed explicitly.

如果复制构造函数不是显式的，但我想知道并实现显式定义而不是猜测正确的转换代码工作正常。

The code works alright if copy constructor is not explicit but I would like to know and implement explicit definition instead of "guessing proper cast".

推荐答案

显式拷贝构造函数意味着拷贝构造函数不会被隐式调用，这是在表达式中发生的： p>

The explicit copy constructor means that the copy constructor will not be called implicitly, which is what happens in the expression:

CustomString s = CustomString("test");

此表达式字面意思是：创建一个临时 CustomString 使用接受 const char * 的构造函数。隐式调用 CustomString 的复制构造函数将该临时变量复制到 s 中。

This expression literally means: create a temporary CustomString using the constructor that takes a const char*. Implicitly call the copy constructor of CustomString to copy from that temporary into s.

现在，如果代码是正确的（即如果复制构造函数不是显式的），编译器将避免创建临时并通过构造 s 直接与字符串文字。

Now, if the code was correct (i.e. if the copy constructor was not explicit), the compiler would avoid the creation of the temporary and elide the copy by constructing s directly with the string literal. But the compiler must still check that the construction can be done and fails there.

您可以显式调用复制构造函数：

You can call the copy constructor explicitly:

CustomString s( CustomString("test") );

但是我建议你避免临时的，只是创建 s 与 const char * ：

But I would recommend that you avoid the temporary altogether and just create s with the const char*:

CustomString s( "test" );

这是编译器会做什么...

Which is what the compiler would do anyway...

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