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]):
- 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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