Little 'byte' of confusion

When we say…

Code:

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&(charArray[index * sizeof(int)]);

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…to retrieve the address to an integer value that stored
inside a byte array, how actually it gets pointed to that
value? Isn’t we are only getting the address of a single byte?

As an example:
if index is 5, doesn’t what we get only the address of
the 5x4=20th byte of the byte array? Please someone explain…
Thanks for reading. Have a really smooth day.
- Kasun

Quote:

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Isn’t we are only getting the address of a single byte?

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Yes. A pointer always points to a single byte since it can only hold a single address. However, the type of the pointer specifies how much memory after the initial address is part of the pointed to value and C++ makes sure that the correct type sizes are used.

Code:

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#include <iostream>
#include <cstdlib>

using namespace std;

int main()
{
    unsigned char ca[2 * sizeof(int)];
    int ia[2] = {98, 97};

    memcpy(ca, ia, 2 * sizeof(int));
    int *p0 = (int*)&ca[1 * sizeof(int)];
    unsigned char *p1 = &ca[1 * sizeof(int)];
    cout<<"Sizeof *p0 "<< sizeof *p0 <<' '<< *p0 <<endl;
    cout<<"Sizeof *p1 "<< sizeof *p1 <<' '<< *p1 <<endl;
}

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Oh.. yes. I see my mistake.
- Thanks