How to declare a pointer to a function?

Well, we assume that you know what does it mean by pointer in C. So how do we create a pointer to an integer in C?
Huh..it is pretty simple..

int * ptrInteger; /*We have put a * operator between int 
                    and ptrInteger to create a pointer.*/

Here ptrInteger is a pointer to integer. If you understand this, then logically we should not have any problem in declaring a pointer to a function 🙂

So let us first see ..how do we declare a function? For example,

int foo(int);

Here foo is a function that returns int and takes one argument of int type. So as a logical guy will think, by putting a * operator between int and foo(int) should create a pointer to a function i.e.

int * foo(int);

But Oops..C operator precedence also plays role here ..so in this case, operator () will take priority over operator *. And the above declaration will mean – a function foo with one argument of int type and return value of int * i.e. integer pointer. So it did something that we didn’t want to do. 🙁

So as a next logical step, we have to bind operator * with foo somehow. And for this, we would change the default precedence of C operators using () operator.

int (*foo)(int);

That’s it. Here * operator is with foo which is a function name. And it did the same that we wanted to do.

So that wasn’t as difficult as we thought earlier!

This article is attributed to GeeksforGeeks.org

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