CODEWITHSUNDEEP
c++cfunction-pointers
zeal
zeal

Reputation: 485

How to pass function-pointer to a function in c++?

Here is three functions such as:-

float Plus    (float a, float b) { return a+b; }
float Minus   (float a, float b) { return a-b; }
float Multiply(float a, float b) { return a*b; }

now there is function which takes pointer to a function as one of the argument:-

void Function_Pointer_func(float a, float b, float (*pt2Func)(float, float))
{
   float result = pt2Func(a, b);    // call using function pointer

   cout << " Result = ";  // display result
   cout << result << endl;
}

and to call the above function "Function_Pointer_func" the function is written below

void Replace()
{ 
   Function_Pointer_func(2, 5, /* pointer to function 'Minus' */ Plus);////   (1)
   Function_Pointer_func(2, 5, /* pointer to function 'Minus' */ &Minus);//// (2)

}

Why does above function works fine as function "Function_Pointer_func" takes function-pointer as argument. And if we replace RHS in line

 float result = pt2Func(a, b);    // call using function pointer

of function "Function_Pointer_func" by (*pt2Func)(a, b);then also it works but for (&pt2Func)(a, b);

it gives an error in VS2008:

" error C2064: term does not evaluate to a function taking 2 arguments "

Now replace in the argument of "float (*pt2Func)(float, float)" in function "Function_Pointer_func" by float (pt2Func)(float, float) then all three 

float result = pt2Func(a, b);    // 
float result = (&pt2Func)(a, b); // 
float result = (*pt2Func)(a, b); //

statement works, why? I hope reason of my discomfort lies in understanding the core understanding of function-pointer. Well, I am not presenting the Q? without any good amount of reading but yes i haven't done any intensive research on this so please feel free to recommend some reading in this regard which will sort out my ambiguity.

Thanks for the help in advance.

Upvotes: 4

Views: 398

Answers (2)

user2288008
user2288008

Reputation:

It's c++ standard.

float Plus(float a, float b);
void Function_Pointer_func(float a, float b, float (*pt2Func)(float, float));

Function_Pointer_func(2, 5, Plus); // (1)
...
float result = pt2Func(a, b); // (2)

(1) is conversion of function to pointer (standard 2003, 4.3):

An lvalue of function type T can be converted to an rvalue of
type “pointer to T.” The result is a pointer to the function

(2) is function call (standard 2003, 5.2.2):

For an ordinary function call, the postfix expression shall be either
an lvalue that refers to a function (in which case the function-to-pointer
standard conversion (4.3) is suppressed on the postfix expression), or it
shall have pointer to function type.

[UPDATE] In detail:

void Replace() { 
   Function_Pointer_func(2, 5, Plus);
   Function_Pointer_func(2, 5, &Minus);
}

Minus is function => &Minus is pointer to function, so no conversion, 3-rd argument of Function_Pointer_func fit perfectly. Plus is a function, so to fit Function_Pointer_func it must be converted to pointer. Standard (1) says that it can be done automatically.

Call cases:

void Function_Pointer_func(float a, float b, float (*pt2Func)(float, float)) {
   float result = pt2Func(a, b); // call by pointer, see (2)
   float result = (*pt2Func)(a, b); // convert pointer to function, so 'normal' call
   float result = (&pt2Func)(a, b); // pointer to function pointer, nope, will not work
}

Upvotes: 2

A Guest
A Guest

Reputation: 29

Functions automatically decay into function pointers. In that context,

  • function_name really means &function_name if not specified.

  • &function_name turns a function into a function pointer.

  • *function_name really means *(function_name), which becomes *(&function_name) per above. * and & "cancel out", so to speak and the resulting function_name decays back into &function_name.

Upvotes: 2

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