赋值运算符的布尔和字符串重载（C ++） [英] Boolean and String Overloads of the Assignment Operator (C++)

问题描述

我定义在赋值运算符的多个重载如下：

foo.h中

class Foo
{
private:
    bool my_bool;
    int my_int;
    std::string my_string;
public:
    Foo& operator= (bool value);
    Foo& operator= (int value);
    Foo& operator= (const std::string& value);
};

Foo.cpp中

Foo.cpp

// Assignment Operators.
Foo& Foo::operator= (bool value) {my_bool = value; return *this;}
Foo& Foo::operator= (int value) {my_int = value; return *this;}
Foo& Foo::operator= (const std::string& value) {my_string = value; return *this;}

这是我的main.cpp中（见标有注释惊喜）

Foo boolFoo;
Foo intFoo;
Foo stringFoo;

// Reassign values via appropriate assignment operator.
boolFoo = true;                // works...assigned as bool
intFoo = 42;                   // works...assigned as int
stringFoo = "i_am_a_string";   // SURPRISE...assigned as bool, not string

std::string s = "i_am_a_string";
stringFoo = s;                 // works...assigned as string

// works...but awkward
stringFoo = static_cast<std::string>("i_am_a_string");

问：谁能告诉我，为什么文字的uncasted字符串在布尔上下文中被评价

Question: Can someone tell me why an uncasted string literal is being evaluated in a boolean context?

推荐答案

C ++标准定义了重载解析规则第13.3章，有你发现：

The C++ standard defines overload resolution rules in chapter 13.3, there you find:

13.3.3.2排名隐式转换序列[over.ics.rank]

² 当比较隐式转换序列的基本形式（如13.3.3.1定义）

² When comparing the basic forms of implicit conversion sequences (as defined in 13.3.3.1)

- 一个标准的转换序列（13.3.3.1.1）比一个用户定义的转换序列或省略号转换序列，以及

— a standard conversion sequence (13.3.3.1.1) is a better conversion sequence than a user-defined conversion sequence or an ellipsis conversion sequence, and

- 用户自定义转换序列（13.3.3.1.2）比一个省略号转换序列（13.3.3.1.3）更好的转换序列

— a user-defined conversion sequence (13.3.3.1.2) is a better conversion sequence than an ellipsis conversion sequence (13.3.3.1.3).

这意味着，编译器将preFER从到布尔或 INT 如果有的话。现在，该标准转换是相关的？在你的情况，这两个是相关的：

This means that the compiler will prefer a standard conversion sequence from the string literal to bool or int if available. Now, which standard conversions are relevant? In your case, these two are relevant:

4.2阵列到指针转换[conv.array]

¹ 左值或类型数组NT·的或必然的T未知的数组的右值可以转换为类型指针T的prvalue。其结果是一个指针数组的第一元素

¹ An lvalue or rvalue of type "array of N T" or "array of unknown bound of T" can be converted to a prvalue of type "pointer to T". The result is a pointer to the first element of the array.

该变换将字符串，它的类型的为const char [N] ，到为const char * 。第二个是：

This conversion turns the string literal, which is of type const char[N], into a const char*. The second one is:

4.12布尔转换[conv.bool]

¹ 算术，无范围枚举，指针或成员指针类型的prvalue可以转换为类型的prvalue 布尔。零值，空指针值或空成员指针值转换为假;任何其他的值转换为真正。类型的prvalue 的std :: nullptr_t 可转换为类型的prvalue 布尔;结果值假。

¹ A prvalue of arithmetic, unscoped enumeration, pointer, or pointer to member type can be converted to a prvalue of type bool. A zero value, null pointer value, or null member pointer value is converted to false; any other value is converted to true. A prvalue of type std::nullptr_t can be converted to a prvalue of type bool; the resulting value is false.

这就是为什么指针被转换为布尔的原因。由于标准转换序列存在，用户自定义转换到的std ::字符串不被使用。

That is the reason why the pointer is converted to bool. Since a standard conversion sequence exists, the user-defined conversion to std::string is not used.

要解决你的问题，我建议你添加其他重载版本，采用为const char * 并使其调用转发到常量性病： ：串放; 超载

To solve your problem, I suggest you add another overloaded version that takes const char* and make it forward the call to the const std::string& overload.

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