转换运算符可以是显式的吗？ [英] Can a cast operator be explicit?

问题描述

当涉及到构造函数时，添加关键字 explicit 可以防止热情的编译器在不是程序员的第一个意图时创建对象。此类机制是否也可用于投放运算符？

When it comes to constructors, adding the keyword explicit prevents an enthusiastic compiler from creating an object when it was not the programmer’s first intention. Is such mechanism available for casting operators too?

struct Foo
{
    operator std::string() const;
};

例如，我想要能够 Foo 转换为 std :: string ，但我不希望这种转换隐式发生。

Here, for instance, I would like to be able to cast Foo into a std::string, but I don’t want such cast to happen implicitly.

推荐答案

是和否。

这取决于您使用的C ++版本。

It depends on which version of C++, you're using.

• C ++ 98和C ++ 03不支持显式类型转换操作符


• 但是C ++ 11。

• C++98 and C++03 do not support explicit type conversion operators
• But C++11 does.

示例，

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 
}

g ++ -std = c ++ 0x ，你将得到这个错误：

Compile it with g++ -std=c++0x, you will get this error:

prog.cpp： 20：错误：从'A'转换为非标量类型'std :: string'请求

prog.cpp:13:20: error: conversion from 'A' to non-scalar type 'std::string' requested

在线演示： http://ideone.com/DJut1

但是一旦你写：

std::string s = static_cast<std::string>(a); //ok - explicit conversion 

错误消失： http://ideone.com/LhuFd

BTW，在C ++ 11中，显式转换如果它转换为 boolean ，则称为上下文转换运算符。此外，如果您想进一步了解隐式和显式转换，请参阅以下主题：

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:

• 隐式VS显式转换

• Implicit VS Explicit Conversion

。

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