C++ function types

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攒了一身酷
攒了一身酷 2020-12-10 01:11

I have a problem understanding function types (they appear e.g. as the Signature template parameter of a std::function):

typedef in         


        
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  • 2020-12-10 01:37

    Here's the relevant paragraph from the Standard. It pretty much speaks for itself.

    8.3.5/10

    A typedef of function type may be used to declare a function but shall not be used to define a function (8.4).

    Example:

    typedef void F();
    F  fv;         // OK: equivalent to void fv();
    F  fv { }      // ill-formed
    void fv() { }  // OK: definition of fv
    

    A typedef of a function type whose declarator includes a cv-qualifier-seq shall be used only to declare the function type for a non-static member function, to declare the function type to which a pointer to member refers, or to declare the top-level function type of another function typedef declaration.

    Example:

    typedef int FIC(int) const;
    FIC f;               // ill-formed: does not declare a member function
    struct S {
      FIC f;             // OK
    };
    FIC S::*pm = &S::f;  // OK
    
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  • 2020-12-10 01:58

    In your case, std_fun_1 and std_fun_2 are identical objects with identical type signatures. They are both std::function<int(int)>, and can both hold function pointers or callable objects of type int(int).

    pf is a pointer to int(int). That is, it serves the same basic purpose as std::function, but without the machinery of that class or support for instances of callable objects.

    Similarly, std_fun_3 and std_fun_4 are identical objects with identical type signatures, and can both hold function pointers or callable objects of type int(int) const.

    Also similarly, pfc is a function pointer of type int(int) const, and can hold pointers to functions of that type, but not instances of callable objects.

    But f and fc are function declarations.

    The line:

    Signature fc;
    

    Is identically equivalent to:

    int fc(int) const;
    

    Which is a declaration for a function named fc of type int(int) const.

    There's nothing strange going on here. You've simply happened upon syntax you probably already understand, from a perspective you're not accustomed to.

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