C++ 强制转换运算符可以是显式的吗？
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Can a cast operator be explicit?
提问by qdii
When it comes to constructors, adding the keyword explicitprevents an enthusiastic compiler from creating an object when it was not the programmer's first intention. Is such mechanism available for casting operators too?
对于构造函数，添加关键字explicit可以防止热心的编译器在不是程序员的本意时创建对象。这种机制也适用于铸造操作员吗？
struct Foo
{
operator std::string() const;
};
Here, for instance, I would like to be able to cast Foointo a std::string, but I?don't want such cast to happen implicitly.
例如，在这里，我希望能够转换Foo为 a std::string，但我不希望这种转换隐式发生。
回答by Nawaz
Yes and No.
是和否。
It depends on which version of C++, you're using.
这取决于您使用的 C++ 版本。
- C++98 and C++03 do not support
explicittype conversion operators - But C++11 does.
- C++98 和 C++03 不支持
explicit类型转换运算符 - 但是 C++11 可以。
Example,
例子，
struct A
{
//implicit conversion to int
operator int() { return 100; }
//explicit conversion to std::string
explicit operator std::string() { return "explicit"; }
};
int main()
{
A a;
int i = a; //ok - implicit conversion
std::string s = a; //error - requires explicit conversion
}
Compile it with g++ -std=c++0x, you will get this error:
用 编译它g++ -std=c++0x，你会得到这个错误：
prog.cpp:13:20: error: conversion from 'A' to non-scalar type 'std::string' requested
prog.cpp:13:20: 错误：请求从“A”转换为非标量类型“std::string”
Online demo : http://ideone.com/DJut1
在线演示：http: //ideone.com/DJut1
But as soon as you write:
但是一旦你写：
std::string s = static_cast<std::string>(a); //ok - explicit conversion
The error goes away : http://ideone.com/LhuFd
错误消失了：http: //ideone.com/LhuFd
BTW, in C++11, the explicit conversion operator is referred to as "contextual conversion operator"if it converts to boolean. Also, if you want to know more about implicit and explicit conversions, read this topic:
顺便说一句，在 C++11 中，如果显式转换运算符转换为boolean ，则它被称为“上下文转换运算符”。此外，如果您想了解有关隐式和显式转换的更多信息，请阅读此主题：
Hope that helps.
希望有帮助。
