I have a function called add which adds two integers as shown below:

int add(int a, int b){
    return a + b;
}

and I wanted to create a pointer to my function named a and call my function using it

printf("The sum of 45, 23: %dn", a(45, 23));

I did it by creating a type for my function with typedef

typedef int (*Aptr) (int, int);

Aptr a = &add;

printf("The sum of 45, 23: %dn", a(45, 23));

The sum of 45, 23: 68

But when I tried to use the type in the typedef statement with my variable, it causes an error and I tried removing the replacement name too but it is throwing an error. If the purpose of typedef is just creating a new name for existing data type why is this not working?

int (*Aptr) (int, int); a = &add;

int (*) (int, int); a = &add;

For both cases:

[Error] 'a' undeclared (first use in this function)
[Note] each undeclared identifier is reported only once for each function it appears in
[Warning] implicit declaration of function 'a' [-Wimplicit-function-declaration]

is there any other way to state the data type of a function to function pointer?