函数的隐式声明sbrk（）

Question

I am trying to create my own malloc function, and I did not finished yet. Here is related part of my code:

mymalloc.h :

#pragma once

#include <stdlib.h>

typedef struct METADATA{
    struct METADATA *next;
    struct METADATA *prev;
    int free;
    size_t size;
}METADATA;

METADATA *metadata;

void *mm_malloc(size_t size);

mymalloc.c

#include "mm_alloc.h"
#include "stdlib.h"
#include "stdio.h"

void *mm_malloc(size_t size) {
    if(size == 0)
        return NULL;

    METADATA *tmp;
    long address;

    if(metadata == NULL){
        sbrk(sizeof(tmp));
        sbrk(size);
        address = sbrk(0);

        return (void *)address;
    }
}

In the sbrk(sizeof(tmp)); part of mymalloc.c , I got "Implicit declaration of function sbrk()" . What I wanted to do ise creating a place for metadata of new block and also a place for the required size. Where am I doing wrong?

Answer 1

Add the line

#include <unistd.h>

At the top of the file 'mymalloc.c' so that the function is declared