#include <stdio.h>
// A normal function with an int parameter
// and void return type
void fun(int a)
{
printf("Value of a is %d\n", a);
}
int main()
{
// fun_ptr is a pointer to function fun()
void (*fun_ptr)(int) = &fun;
/* The above line is equivalent of following two
void (*fun_ptr)(int);
fun_ptr = &fun;
*/
// Invoking fun() using fun_ptr
(*fun_ptr)(10);
return 0;
}
Value of a is 10
Why do we need an extra bracket around function pointers like fun_ptr in above example? If we remove bracket, then the expression ?void (*fun_ptr)(int)? becomes ?void *fun_ptr(int)? which is declaration of a function that returns void pointer.
First understand the algorithm carefully. Then study the program line-by-line and compare it with the output. Finally, review the explanation section to strengthen your logic and programming understanding.
Rewrite the program without looking at the code. Modify values, conditions or logic and run it again. This helps improve confidence and strengthens coding skills much faster.