#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 read the algorithm, then study the program code line by line. After that, compare the code with the output and finally go through the explanation. This approach helps learners understand both the logic and the implementation properly.
After understanding this example, try to rewrite the same program without looking at the code. Then change some values or logic and run it again. This helps improve confidence and keeps learners engaged on the page for longer.