How is the array stored in memory? [closed]

Answer 1

Since I can't add comments just yet, here's my two cents in an answer:

If you only want to know if the memory is on the stack or heap, read the other answers, they are much more informed than me.

If you want to know exactly where the values are, you can always print the address:

printf("address at a[0] = %p\n", (void *)&a[0]);
printf("address at p[0] = %p\n", (void *)&p[0]);

where you will notice the same answer. But, then look at

printf("address at a[1] = %p\n", (void *)&a[1]);
printf("address at p[1] = %p\n", (void *)&p[1]);

Which is a fun little exercise. Just for fun, run the following code and see what you get:

 p[2] = 'a';
 printf("a[0] is %d\n", a[0]);
 printf("a[1] is %d\n", a[1]);
 printf("p[2] is %d\n", p[2]);
 putchar(p[2]);

Answer 2

A statically-created array will be on the stack or in the .data/.bss sections of your binary. A dynamically-created array (with new or malloc) will be allocated on the heap.

Answer 3

An array stores its elements in contiguous memory locations.
If You created the array locally it will be on stack. Where the elements are stored depends on the storage specification.
For Example:
An array declared globally or statically would have different storage specification from an array declared locally. Technically, the where part is implementation defined but usually implementations would use similar usage patterns.

• A local array will be (usually) created on stack while
• A global or static array will be (usually) created on bss/data segments and
• A dynamically created array will be created on heap.

Answer 4

First, you pointer has to be of type int. An array is just a group of integer, saved in the memory as single integer, but in one row. A integer has 4-Byte in the memory, so you can access each value of your array by increasing your pointer by 4.

int *p = (int*)a;
for(i = 0; i < 5; i++) {
    p += 4;
    printf("%d ", *p);
}