Function variable instead of pointer to function

Question

There is this code:

void a() { }

// function parameter first defined as function, then as pointer to function
void g(void f() = a, void (*d)() = a) {
}

int main(){
   void (*z)() = a; // works ok
   //void z() = a; error: illegal initializer (only variables can be initialized)
   return 0;
}

In function parameter list you can define function parameter as a function or as pointer to function (see function g parameter list). However in main function only pointer to function version seems to work. Is there some way to define a function variable (not a pointer to function) in block scope (like the void f() = a variable in g function)?

Kerrek SB · Accepted Answer · 2013-05-18 15:30:07Z

2

You cannot have variables of function type in C++. You can either have pointers or references to functions, though:

typedef void (funtype)();

void call(funtype & f) { f(); } // OK, funtype & == void (&)()
void call(funtype * g) { g(); } // Also OK, funtype * == void (*)()

In some sense, you always call a function through a function pointer: An expression whose value is a function immediately decays to the corresponding function pointer. So in the above examples, f(), (*f)(), (**f)() and so forth are all valid ways of calling the function, and likewise for g. But passing by reference has the advantage that you can still get the function pointer explicitly by saying &f.

answered May 18, 2013 at 15:30

Kerrek SB

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Comments

Joseph Mansfield · Accepted Answer · 2013-05-18 15:31:43Z

0

The argument f of your function g is actually of function pointer type. Any function type argument is transformed to a function pointer type:

After determining the type of each parameter, any parameter of type “array of T” or “function returning T” is adjusted to be “pointer to T” or “pointer to function returning T,” respectively.

However, that doesn't happen for a normal variable declaration. void z() = a; declares a function and then attempts to initialise it. You can't do this because a function is not a variable.

answered May 18, 2013 at 15:31

Joseph Mansfield

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Comments

Mats Petersson · Accepted Answer · 2013-05-18 15:31:48Z

0

There is no such thing as a "function variable", it must be a pointer or a reference to a function. (Other languages probably simply hide the fact that the variable is a pointer or reference).

answered May 18, 2013 at 15:31

Mats Petersson

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Pete Becker · Accepted Answer · 2013-05-18 15:31:55Z

0

In void g(void f() = a), f is not a function type; it's a pointer to function. (Cf. void h(int x[3]), where x is a pointer to int). In void f() = a; f is, indeed, a function type and you cannot create a variable of a function type.

answered May 18, 2013 at 15:31

Pete Becker

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