What does “typedef void *(*Something)(unsigned int)” mean

Question

I am wondering, what does below mean in.h, it has

 typedef void *(*some_name)(unsigned int);

And then in.c

some_name rt;
some_name State= 0;
unsigned int t = 1;
rt = (some_name) State(t);

Answer 1

It creates an alias some_name for a pointer to a function with a return type void* and a single unsigned int parameter. An example:

typedef void *(*my_alloc_type)(unsigned int);

void *my_alloc(unsigned int size)
{
    return malloc(size);
}

int main(int argc, char *argv[])
{
    my_alloc_type allocator = my_alloc;
    void *p = allocator(100);
    free(p);
    return 0;
}

Answer 2

This typedef

typedef void *(*some_name)(unsigned int);

introduces an alias for pointer to a function of the type void *( unsigned int ) that is that has the return type void * and one parameter of the type unsigned int .

As for this code snippet

some_name rt;
some_name State= 0;
unsigned int t = 1;
rt = (some_name) State(t);

then it does not make a sense.

Answer 3

typedef void *(*)(unsigned int);
                ^^^^^^^^^

somename is

typedef void *some_name(unsigned int);
              ^^         ^

a pointer

typedef void *(*some_name);
                          ^^^^^^^^^^^^^^

to a function that accepts an unsigned int

typedef (*some_name)(unsigned int);
        ^^^^^^

and returns a pointer to void

 void *(*some_name)(unsigned int);
^^^^^^^

but for types.

---

So "the type somename is for functions that accept unsigned and return pointer to void"

#include <stdlib.h>
typedef void *(*some_name)(unsigned int);
void *foo(unsigned t) {
    if (t == 0) return NULL;
    return malloc(t);
}
int main(void) {
    somename fxptr = foo;
    void *bar = fxptr(42);
    free(bar);
}