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.
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 integersizeof(*A) => 4 because *A means value at starting address and that needs 4 bytes to storesizeof(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.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.
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.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.
sizeof(A), you get the size ofA. 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.If you assume that A is always the same as A+0no I assumedAas&A[0]&A[0]is the same asA+0in this case.