malloc 如何区分我分配一个 long long 整数和一个由两个整数组成的数组？

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 ?

Answer 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.

Answer 2

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.