C function pointer syntax

Solution 1:

I explain this in my answer to Why was the C syntax for arrays, pointers, and functions designed this way?, and it basically comes down to:

the language authors preferred to make the syntax variable-centric rather than type-centric. That is, they wanted a programmer to look at the declaration and think "if I write the expression *func(arg), that'll result in an int; if I write *arg[N] I'll have a float" rather than "func must be a pointer to a function taking this and returning that".

The C entry on Wikipedia claims that:

Ritchie's idea was to declare identifiers in contexts resembling their use: "declaration reflects use".

...citing p122 of K&R2.

Solution 2:

This structure reflects how a normal function is declared (and used).

Consider a normal function definition:

int foo (int bar, int baz, int quux);

Now consider defining a function pointer to a function of the same signature:

int (*foo) (int, int, int);

Notice how the two structures mirror each other? That makes *foo much easier to identify as a function pointer rather than as something else.

Solution 3:

If you're dealing with a function (not a pointer to one), the name is in the middle too. It goes like: return-type function-name "(" argument-list ")" .... For example, in int foo(int), int is the return type, foo the name and int the argument list.

A pointer to a function works pretty much the same way -- return type, then name, then argument list. In this case, we have to add a * to make it a pointer, and (since the * for a pointer is prefix) a pair of parentheses to bind the * to the name instead of the return type. For example, int *foo(int) would mean a function named foo that takes an int parameter and returns a pointer to an int. To get the * bound to foo instead, we need parentheses, giving int (*foo)(int).

This gets particularly ugly when you need an array of pointers to functions. In such a case, most people find it easiest to use a typedef for the pointer type, then create an array of that type:

typedef int (*fptr)(int);

fptr array[10];

Solution 4:

I had seen at some places function pointers declared as

int (*foo) (int a, int b);

and at some places a and b are not mentioned and both still works.

so

int (*foo) (int, int)

is also correct.

A very simple way that I found to remember is as mentioned below:

Suppose function is declared as:

int function (int a , int b);

Step1: Simply put function in parentheses:

int (function) (int a , int b);

Step2: Place a * in front of function name and change the name:

int (*funcPntr) (int a , int b);

PS: I am not following proper coding guidelines for naming convention etc. in this answer.