How to declare this function...

Declare a pointer to a function which takes array of pointer to a function as an argument and return the pointer to a function.....
how to declare this function....

You really ought to do your own homework...

QuantumPete

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how to declare this function

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use typedef

Code:

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typedef int (* fpointer)(int,char*);

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and use the new type as you need

Code:

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fpointer func(fpointer f[], size_t arr_size);

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Thanks vart.. I am not able to understand can you explain me in detail vart....

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Originally Posted by sreeramu

Thanks vart.. I am not able to understand can you explain me in detail vart....

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What exactly you do not understand? There are only 2 lines of code here...

Do you know what typedef is?

I have used this example in the past for such posts. So this help you as well in explaining an example of using a function pointer and how to declare it, outside of main.


http://www.cplusplus.com/doc/tutorial/pointers.html

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int addition (int a, int b) {
  return (a+b); }

int subtraction (int a, int b) {
  return (a-b);}

int operation (int x, int y, int (*functocall)(int,int))
{
  int g;
  g = (*functocall)(x,y);  //A pointer to a function type 4.12 and 1.34 of comp.lang.c FAQ.
  return (g);
}

int main ()
{
  int m,n;
  int (*minus)(int,int) = subtraction;

  m = operation (7, 5, addition);
  n = operation (20, m, minus);
  printf("%d\n", n);
  return 0;
}

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Note:

A function call actually requires a function pointer value before the
parentheses. In a typical "direct" function call, this pointer value
is obtained from the name of the function.
A function name (more generally, any expression of function type) is
implicitly converted to a pointer to the function in most contexts.
(The exceptions are when it's the operand of a sizeof operator, which
is illegal rather than yielding a pointer size, and when it's the
operand of a unary "&", which yields the address of the function.)

Code:

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g = (*functocall)(x,y);

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could be just

Code:

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g = functocall(x,y);

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