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

我正在定义赋值运算符的多个重载，如下所示:

I am defining multiple overloads of the assignment operator as follows:

Foo.h

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

// 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(请参阅标记为 SURPRISE 的评论):

And here's my main.cpp (see the comment marked SURPRISE):

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");

问题:有人能告诉我为什么要在布尔上下文中评估未转换的字符串文字吗?

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

推荐答案

C++ 标准在第 13.3 章中定义了重载解析规则，你可以找到:

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

13.3.3.2 对隐式转换序列进行排名[over.ics.rank]

²比较隐式转换序列的基本形式时(定义见13.3.3.1)

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

——标准转换序列 (13.3.3.1.1) 是比用户定义的转换序列或省略号转换序列更好的转换序列，并且

— 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

- 用户定义的转换序列 (13.3.3.1.2) 是比省略号转换序列 (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).

这意味着编译器将首选从字符串文字到 bool 或 int 的标准转换序列(如果可用).现在，哪些标准转换是相关的?在您的情况下，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 数组到指针的转换[conv.array]

¹ N T 的数组"或T 的未知边界数组"类型的左值或右值可以转换为指向 T 的指针"类型的纯右值.结果是指向数组第一个元素的指针.

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.

此转换将类型为 const char[N] 的字符串文字转换为 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 算术纯右值、无作用域枚举、指针或指向成员类型的指针可以转换为 bool 类型的纯右值.零值、空指针值或空成员指针值转换为false；任何其他值都将转换为 true.std::nullptr_t 类型的纯右值可以转换为 bool 类型的纯右值；结果值为 false.

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.

这就是指针转换为bool的原因.由于存在标准转换序列，因此不使用用户定义的到 std::string 的转换.

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.

为了解决您的问题，我建议您添加另一个采用 const char* 的重载版本，并将调用转发到 const std::string& 重载.

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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