C++ 显式复制构造函数
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/11480545/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Explicit copy constructor
提问by abumusamq
I have extended std::string to fulfil my needs of having to write custom function build into string class called CustomString
我扩展了 std::string 以满足我必须将自定义函数构建写入名为CustomString 的字符串类的需要
I have defined constructors:
我已经定义了构造函数：
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);
... };
In the third constructor (copy constructor) and assignment operator, whose definition is:
在第三个构造函数（复制构造函数）和赋值运算符中，其定义为：
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;
}
First since this is an "explicit"; meaning an explicit cast is needed to assign to another CustomString object; it's complaining about the assignment.
首先，因为这是一个“显式”；这意味着需要显式转换才能分配给另一个 CustomString 对象；它在抱怨这个任务。
CustomString s = CustomString("test");
I am not sure where exactly is casting needed explicitly.
我不确定在哪里明确需要强制转换。
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".
如果复制构造函数不是显式的，代码可以正常工作，但我想知道并实现显式定义，而不是“猜测正确的转换”。
回答by David Rodríguez - dribeas
The explicit copy constructor means that the copy constructor will not be called implicitly, which is what happens in the expression:
显式复制构造函数意味着不会隐式调用复制构造函数，这就是表达式中发生的情况：
CustomString s = CustomString("test");
This expression literally means: create a temporary CustomStringusing the constructor that takes a const char*. Implicitly call the copy constructor of CustomStringto copy from that temporary into s.
这个表达式的字面意思是：CustomString使用带有const char*. 隐式调用 的复制构造函数CustomString从那个临时复制到s.
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 sdirectly with the string literal. But the compiler must still check that the construction can be done and fails there.
现在，如果代码是正确的（即，如果复制构造函数不是显式的），编译器将避免临时的创建并通过s直接用字符串字面量构造来省略副本。但是编译器仍然必须检查构造是否可以完成并在那里失败。
You can call the copy constructor explicitly:
您可以显式调用复制构造函数：
CustomString s( CustomString("test") );
But I would recommend that you avoid the temporary altogether and just create swith the const char*:
不过，我建议你完全避免暂时的，只是创建s与const char*：
CustomString s( "test" );
Which is what the compiler would do anyway...
这就是编译器无论如何都会做的......
回答by ForEveR
Deriving from std::string is not safe, as std::string has no virtual destructor. As to your question - your copy constructors should not be explicit, to allow for such usage as:
从 std::string 派生是不安全的，因为 std::string 没有虚拟析构函数。至于你的问题 - 你的复制构造函数不应该是明确的，以允许这样的用法：
CustomString s = "test";
Also I have no idea why you would want to declare a copy-constructor as explicit, as it is not needed. An explicit copy-constructor will work only if you declare your CustomString object as:
此外，我不知道您为什么要将复制构造函数声明为显式，因为它不需要。只有将 CustomString 对象声明为：
CustomString s(CustomString("test"));
