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cpointersfunction-pointers
Kamie Xeng
Kamie Xeng

Reputation: 37

define a pointer variable to the function in C program

If I want to define a pointer variable p to point to the function foo() defined as below, what should be the exact type of p?

int *foo(void *arg)
{
    ...
}

Upvotes: 2

Views: 136

Answers (5)

user3860869
user3860869

Reputation: 121

Are you talking about pointer to function?

First define a pointer to function that takes void argumenta and return int

typedef   int (*Ptrtofun)(void);

Now safely can point this to your function.

Ptrtofun p;
p = &foo(void);

Now you have a ptr p to function and can easily use it ..

Upvotes: 1

rici
rici

Reputation: 241691

Bearing in mind the comment of Kyrylo Polezhaiev, the type of p is

int*(*)(void*)

To declare p, you need to insert the name at the correct point in the type:

int*(*p)(void*);

but it is generally more readable to use an alias:

typedef int*(func_t)(void*);
func_t* p;

Upvotes: 3

Vlad from Moscow
Vlad from Moscow

Reputation: 310940

To make it more easy you could for the function declaration

int *foo(void *arg);

define an alias named as for example Tfoo that looks the same way as the function declaration itself

typedef int *Tfoo(void *);

After that to declare a function pointer to the function is very easy. Just write

Tfoo *foo_ptr = foo;

Otherwise you could write

typedef int * (* TFoo_ptr )(void *);

Tfoo_ptr foo_ptr = foo;

Or you could write a declaration of the function pointer directly without using the typedef/

int * (* foo_ptr )(void *) = foo;

Upvotes: 1

Sourav Ghosh
Sourav Ghosh

Reputation: 134286

You need to have the pointer as the pointer to a function, returning an int *, accepting a void* argument. You can make it like

int * (*p) (void *);

and then, you can use

p = foo;

Upvotes: 4

Iharob Al Asimi
Iharob Al Asimi

Reputation: 53006

It should be

typedef int *(*funtion_foo_type)(void *);

Upvotes: 4

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