条件运算符无法解析重载的成员函数指针 [英] Conditional operator can't resolve overloaded member function pointers

问题描述

我在处理C ++中重载成员函数的指针时遇到了一个小问题。以下代码可以正常编译：

I'm having a minor issue dealing with pointers to overloaded member functions in C++. The following code compiles fine:

class Foo {
public:
    float X() const;
    void X(const float x);
    float Y() const;
    void Y(const float y);
};

void (Foo::*func)(const float) = &Foo::X;

但这不会编译（编译器抱怨重载是模棱两可的）：

But this doesn't compile (the compiler complains that the overloads are ambiguous):

void (Foo::*func)(const float) = (someCondition ? &Foo::X : &Foo::Y);

大概这与编译器将条件运算符的返回值与函数指针类型？我可以解决这个问题，但是我很想知道规范说明所有这些事情应该起作用，因为这似乎有点不直观，并且是否有解决此问题的方法而不会退回到5行if-then-else

Presumably this is something to do with the compiler sorting out the return value of the conditional operator separately from the function pointer type? I can work around it, but I'm interested to know how the spec says all this is supposed to work since it seems a little unintuitive and if there's some way to work around it without falling back to 5 lines of if-then-else.

我正在使用MSVC ++，如果有什么区别。

I'm using MSVC++, if that makes any difference.

谢谢！

推荐答案

从13.4 / 1节开始（重载函数的地址，[over.over]）：

From section 13.4/1 ("Address of overloaded function," [over.over]):

在某些情况下，使用不带参数的重载函数名可以解析为函数，重载集中特定函数的函数指针或成员函数指针。在这种情况下，功能模板名称被认为是命名一组重载功能。所选函数的类型与上下文中所需的目标类型相匹配。目标可以是

A use of an overloaded function name without arguments is resolved in certain contexts to a function, a pointer to function or pointer to member function for a specific function from the overload set. A function template name is considered to name a set of overloaded functions in such contexts. The function selected is the one whose type matches the target type required in the context. The target can be

• 正在初始化的对象或引用（8.5、8.5.3），


• 分配的左侧（5.17），


• 函数的参数（5.2.2），


• 用户的参数-定义的运算符（13.5），


• 函数，运算符或转换的返回值（6.6.3）或


• 显式类型转换（5.2.3、5.2.9、5.4）。

• an object or reference being initialized (8.5, 8.5.3),
• the left side of an assignment (5.17),
• a parameter of a function (5.2.2),
• a parameter of a user-defined operator (13.5),
• the return value of a function, operator function, or conversion (6.6.3), or
• an explicit type conversion (5.2.3, 5.2.9, 5.4).

重载函数名称可以以& 运算符。在没有列出的上下文中，没有参数的情况下，不能使用重载的函数名。 [注意：忽略重载的函数名周围的任何多余的括号集（5.1）。 ]

The overload function name can be preceded by the & operator. An overloaded function name shall not be used without arguments in contexts other than those listed. [Note: any redundant set of parentheses surrounding the overloaded function name is ignored (5.1). ]

您希望从上面的列表中选择的 target 是第一个对象正在初始化。但是有一种条件运算符，条件运算符是根据操作数而不是任何目标类型来确定其类型的。

The target you were hoping would be selected from the above list was the first one, an object being initialized. But there's a conditional operator in the way, and conditional operators determine their types from their operands, not from any target type.

由于显式类型转换已包含在目标，您可以分别在条件表达式中类型转换每个成员指针表达式。我先做一个typedef：

Since explicit type conversions are included in the list of targets, you can type-cast each member-pointer expression in the conditional expression separately. I'd make a typedef first:

typedef void (Foo::* float_func)(const float);
float_func func = (someCondition ? float_func(&Foo::X) : float_func(&Foo::Y));

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