指向const成员函数typedef的指针 [英] pointer to const member function typedef

问题描述

我知道可以单独创建一个成员函数的指针，像这样

I know it's possible to separate to create a pointer to member function like this

struct K { void func() {} };
typedef void FuncType();
typedef FuncType K::* MemFuncType;
MemFuncType pF = &K::func;

有类似的方法来构造一个指向const函数的指针吗？我试着添加const在各个地方没有成功。我已经玩过gcc一些，如果你在像

Is there similar way to construct a pointer to a const function? I've tried adding const in various places with no success. I've played around with gcc some and if you do template deduction on something like

template <typename Sig, typename Klass>
void deduce(Sig Klass::*);

它将显示Sig作为一个函数签名与const刚刚粘在末端。如果在代码中这样做，它会抱怨你不能有一个函数类型的限定符。

It will show Sig with as a function signature with const just tacked on the end. If to do this in code it will complain that you can't have qualifiers on a function type. Seems like it should be possible somehow because the deduction works.

推荐答案

您希望这样：

typedef void (K::*MemFuncType)() const;

如果您仍然要 MemFuncType FuncType ，您需要更改 FuncType ：

If you want to still base MemFuncType on FuncType, you need to change FuncType:

typedef void FuncType() const;
typedef FuncType K::* MemFuncType;

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