How does malloc distinguish between me allocating a long long integer and an array of two integers?

Question

in my OS long long int is : 8 bytes . int is 4 bytes .

int *p = malloc(4);

this code allocate 4 bytes for a variable of type integer on the heap .

int *p = malloc(8);

will this allocate a long long integer like 'one variable' or two items on an array .

how can i allocate an integer of 8 bytes long ? how can i allocate an array containing 2 items ?

`malloc()` doesn't know anything about the type of data, it just allocates memory. — Barmar, Oct 30 '20 at 21:54
`malloc()` does not care about data type sizes. It's up to you to do the math when allocating memory of size * your data type size. — Irelia, Oct 30 '20 at 21:56
Related to: https://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc (first comment to the question and accepted answer) — Jose, Oct 30 '20 at 21:59

Barmar · Accepted Answer · 2020-10-30T22:14:54.290

1

malloc() just allocates raw memory, whether it's treated as an array or a single variable is determined by how the memory is used in the caller.

int *p = malloc(2 * sizeof(int));

treats the memory as an array of 2 int. You can then do:

p[0] = 1;
p[1] = 2;

and it will write two int into the memory.

long int *p = malloc(sizeof(long int));

treats the memory as a single long int (or, equivalently, an array of 1 long int). Then you can do:

*p = 12345678;

and it will write that long integer into the memory.

Both of them will allocate 8 bytes of memory on a system where int is 4 bytes and long int is 8 bytes.

edited Oct 30 '20 at 22:14

answered Oct 30 '20 at 21:58

Barmar

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so i need to specify pointer type(long int *p=) so the program know that i want a long integer and not array of 2 integers – soldier captain Oct 30 '20 at 22:01
Yes. When you use `p` it uses its type to know how to interpret the memory it points to. – Barmar Oct 30 '20 at 22:03
That's the reason why you have to specify the type of the pointer. – Barmar Oct 30 '20 at 22:03
so what will happen if i do int *p=malloc(5) – soldier captain Oct 30 '20 at 22:06
The first 4 bytes of the allocation will be used when you dereference the pointer, the extra byte will be ignored. – Barmar Oct 30 '20 at 22:07

tadman · Answer 2 · 2020-10-30T22:03:34.770

If you want an 8 byte integer, it's called a long long or int64_t/int64_t:

If this is about C, then you use malloc:

int64_t* p = malloc(sizeof(int64_t));

If this is about C++, then you use new:

int64_t* p = new int64_t;

With your original code you're getting an allocation of arbitrary size mapped to a pointer of a fixed size type, where int is typically 4 bytes, or in other words, you have room for int[2].

Note: For portability reasons it's always best to express your allocations in terms of base types, not just abstract numbers. malloc(8) may allocate memory for 2 x int, or it may not, that depends on what sizeof(int) is. malloc(sizeof(int) * 2) always works correctly.

How does malloc distinguish between me allocating a long long integer and an array of two integers?

2 Answers2