Understanding memory allocation behaviour for arrays using malloc

Question

I have this simple code:

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

int main(){ 

int (*array)[2] = malloc(sizeof(int)*2);
    printf("%p\n",array[0]); //0x13c606700
    printf("%p\n",array[0]+1); //0x13c606704
    printf("%p", array[1]); //0x13c606708
}

I'm using malloc to allocate memory for an integer array of 2 elements. This returns a pointer for this said array. However, I don't understand why array[0]+1 and array[1] are yielding different addresses. array[0]+1 prints the address at array[0] + 4 which is expected since an integer is of size 4 bytes. But this is not the same for array[1]. Why is this happening? Wouldn't intuition suggest that using malloc on a fixed-size array would enable the programmer to refer to the memory allocated using array notation (i.e array[i])?

Your declaration is almost certainly not what you want. You have declared a pointer to an array of two int. This is normally used for a two-dimensional array. If you just want a dynamically allocated array of two int, the declaration should be int *array. You can use what you have for a one-dimensional array, but it's unusual, and your references would need to look like (*array)[i] which isn't what you're doing. — Tom Karzes
– Tom Karzes, Commented Oct 28, 2021 at 14:49
An "integer array of 2 elements" is int *array (to be allocated) or int array[2]. For the first you need int *array = malloc(sizeof *array * 2); but then the data at array[0] and array[1] is indeterminate. — Weather Vane
– Weather Vane, Commented Oct 28, 2021 at 14:50

Ted Lyngmo · Accepted Answer · 2021-10-28 14:52:19Z

2

array[0] is an int[2]. When passed to the function it decays into a pointer to the first element, an int, which is 4 bytes on your system.

array[0] + 1 adds the sizeof(int) to the pointer.

array[1] is the next int[2] (out of bounds). That's the sizeof(int) times two.

answered Oct 28, 2021 at 14:52

Ted Lyngmo

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2 Comments

Nizar Over a year ago

If I may ask you (and apologise for my lack of knowledge). By using (*array)[2], did I allocate memory for 2 x 1d arrays of sizeof(int)*2 ? Meaning 2 arrays of 8 bytes?

Ted Lyngmo Over a year ago

@Nizar No, you only allocated space for two ints, so that's one int[2].

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