从函数返回指针

Question

I am new in C and literally trying to return pointer from my function to the pointer variable and have this "[Warning] assignment makes pointer from integer without a cast" no idea why compiler defines it as an int. Can't declare my function before main as well, it throws this "undefined reference to `free_block'".

#include <stdio.h>
#include <stdlib.h>
    struct block{
    int num;
};

int main(int argc, char *argv[]) {
    struct block *b;
    b = free_block();

    struct block *free_block(){
        struct block *b = NULL;
        return b;
    }
    return 0;
}

Thank you

Yea, my fault I know not too much about c syntax and had no idea about nested functions, soz. But what could be wrong in this case: I am trying to make my own memory allocator without using malloc or calloc functions. In my code I have the same Warning on the line with pointer = free_space_get(size); , here I have no more nested func(), my methods defined before main(), but still have no idea do I have to declare my functions or no, coz in the answer given to me it worked fine as soon as functions were defined before the main().

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct header{
        size_t size;
        struct header *next;
        unsigned int free;
    };
void *m_alloc(size_t  size){
        size_t total_size;
        void *block;
        struct header *pointer;
        if(!size)  
            return NULL;
        pointer = free_space_get(size);
        if(pointer){
            pointer->free = 0;
            return (void*)(pointer + 1);
        }

    }
struct header *get_free_space(size_t size){
        struct header *b = NULL;
        return b;
    }

int main() {
    return 0;
}

Answer 1

Your code can be re-written as

#include <stdio.h>
#include <stdlib.h>

struct block{
    int num;
};

struct block *free_block(){
        struct block *b = NULL;
        return b;
}

int main(int argc, char *argv[]) {

    struct block *b;
    b = free_block();

    if(b == NULL) // Checking whether pointer is returned
        printf("\n Recieved NULL \n");

    return 0;
}