Array, pointer and Sizeof relation

Question

I have learned that we can use the address of elements of an array at index i as &A[i] or simply as A+i and to get the value A[i] or *(A+i) where A is an array and i denotes the Index. Let's say int A[]={1,2,3};

When I use sizeof(*A) I get value 4, now when I use sizeof(A) I must get the size of address value of the first element why I get the size of the whole array as 12. I am a beginner and confused, please guide.

"when I use sizeof(A) I must get the size of address value" no, you should get the size of A. Since A is an array, you are getting the size of the array. — n. m. could be an AI, Jan 28 '18 at 17:14
Since &A[i] is same as (A+i) , so if I do sizeof(A) , I must get the capacity of address pointing to array's first element , hope you understand — anime, Jan 28 '18 at 17:16
The rules of the language are such that if you say `sizeof(A)`, you get the size of `A`. Not the size of `(A+0)` and not the size of `&*A`. If you assume that A is always the same as A+0 etc. then you are mistaken. — n. m. could be an AI, Jan 28 '18 at 17:30
`If you assume that A is always the same as A+0` no I assumed `A` as `&A[0]` — anime, Jan 28 '18 at 17:32

score 2 · Accepted Answer · answered Jan 28 '18 at 17:27

int A[]={1,2,3};

A is an array and array is not a pointer. when you are printing sizeof(*A) it prints 4 because *A means first element of array which size is 4.

sizeof(A) will results not result in 4 because A is not a pointer. A is an array and array means collection of elements and each elements needs 4 bytes.

A[i] == *(A+i)

sizeof(A[i]) => 4 because A[0] is an integer
sizeof(*A) => 4 because *A means value at starting address and that needs 4 bytes to store
sizeof(A) => 12 bytes, not 4 because A is not a pointer.
sizeof(&A[0]) => 4 bytes because &A[0] yields in address and size of any address will be 4 bytes.

score 0 · Answer 2 · answered Jan 28 '18 at 17:16

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When you call sizeof(A) you get the size of the array as n.m said. But If you initialize your array as this:

    int *A = new int[3];
    A[0] = 1;
    A[1] = 2;
    A[2] = 3;
    cout << sizeof(A) << endl;

you will get 4.

answered Jan 28 '18 at 17:16

wdc

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Since `A` is equal to `&A[0]` for an array, so if I do `printf("%d",A)` I will get the address of first value in my case, so why does `sizeof(A)` don't function in same manner and tells me the sizeof address value pointing to first element of A. – anime Jan 28 '18 at 17:19

sjsam · Answer 3 · 2018-01-28T17:46:57.590

When I use sizeof(*A) I get value 4

This is nothing but size of first element in the array ie sizeof(*A) is the same as sizeof(*(&A[0])).

now when I use sizeof(A) I must get the size of address value of the first element why I get the size of the whole array as 12

You have declared A as

int A[]={1,2,3};

Now, ask yourself a simple question - What is A ? Never settle with anything other than Array for an answer. In short you get the size of entire array in bytes. Here you have 3 elements which takes 4 bytes each.

Added

I must get the size of address value

That should be sizeof(&A) - which is the size of the address of A - which is usually 8 bytes in many systems.

Array, pointer and Sizeof relation

3 Answers3