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Home/ Questions/Q 845579
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T06:28:30+00:00 2026-05-15T06:28:30+00:00

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

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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;

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::*);

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.

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  1. Editorial Team
    Editorial Team
    2026-05-15T06:28:31+00:00Added an answer on May 15, 2026 at 6:28 am

    You want this:

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

    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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