pointers to functions.

[vivekshivu]

#include <stdio.h>
 
int foo(int n)
{printf("Foo got %d\n", n); return(0);
}
 
int bar(int n)
{
printf("Bar got %d\n", n); return(0);
}
 
int main(int argc, char *argv[])
{  
int m = 0;
int (*(fun_array[3]))();
int i = 1;
int ii = i/++i;
printf("\ni/++i = %d, ",ii);
fun_array[1] = foo; 
fun_array[2] = bar; 
(fun_array[++m])(++m);    
}
 

In the above program what does this statement implies
(*(fun_array[3]))(); ?

Is it an array of 3 elements that can store addresss of 3 functions ?
]
And i'm getting 3 errors in it, how to slove them ?
i'm using turboC++ compiler.

[sicarie]

#include <stdio.h>
 
int foo(int n)
{ 
   printf("Foo got %d\n", n); 
   return(0);
}
 
int bar(int n)
{
   printf("Bar got %d\n", n); 
   return(0);
}
 
int main(int argc, char *argv[])
{  
   int m = 0;
   int (*(fun_array[3]))();
   int i = 1;
   int ii = i/++i;
   printf("\ni/++i = %d, ",ii);
   fun_array[1] = foo; 
   fun_array[2] = bar; 
   (fun_array[++m])(++m);    
}
 

In the above program what does this statement implies
(*(fun_array[3]))(); ?

Is it an array of 3 elements that can store addresss of 3 functions ?
]
And i'm getting 3 errors in it, how to slove them ?
i'm using turboC++ compiler.

That depends on what the errors are, let's see them. I'm guessing at least one of them deals with the line you are talking about.

PS - you do know that you have no implementation of fun_array - whatever it is. I wouldn't be surprised if that was an obscure way to declare an array, but I do not believe it is a function.