Thread: Little 'byte' of confusion

  1. #1
    Registered User
    Join Date
    Jan 2004
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    22

    Little 'byte' of confusion

    When we say…

    Code:
    &(charArray[index * sizeof(int)]);
    …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

  2. #2
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    Feb 2004
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    46
    Isn’t we are only getting the address of a single byte?
    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:
    #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;
    }

  3. #3
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    22
    Oh.. yes. I see my mistake.
    - Thanks

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