Implement aligned_malloc using malloc in C [duplicate]

Question

Question: why do we declare p2 as void **? why not p2*?

we are returning p2, but our return function type is void *. This doesn't make any sense. Compiler will say unmatch return type.

void *aligned_malloc(size_t required_bytes, size_t alignment) {
    void *p1;
    void **p2;
    int offset=alignment-1+sizeof（void*);
    p1 = malloc(required_bytes + offset);               // the line you are missing
    p2=(void**)(((size_t)(p1)+offset)&~(alignment-1));  //line 5
    p2[-1]=p1; //line 6
    return p2;
}

I think Maria got the function from the above link. But she is asking a different question though. — Anni_housie
– Anni_housie, Commented Nov 19, 2016 at 8:33

molbdnilo · Accepted Answer · 2016-11-19 08:38:26Z

0

void** can be implicitly converted to void*, so there shouldn't be a type problem.

The reason it's declared void** is to make it convenient to store the allocated pointer just in front of it.

It works like this code, which uses another variable:

void *aligned_malloc(size_t required_bytes, size_t alignment) {
    void *p1;
    void *p2;
    void **p3;
    int offset=alignment-1+sizeof（void*);
    p1 = malloc(required_bytes + offset);
    p2= (void*)(((size_t)(p1)+offset)&~(alignment-1));
    p3 = (void**) p2; 
    p3[-1]=p1;
    return p2;
}

answered Nov 19, 2016 at 8:38

molbdnilo

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1 Comment

Maria Over a year ago

so basically, void ** is to make it become array. And the reason to make it become array is because we need to store the allocated pointer?

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