Can someone please explain the output of this code?

Discussion in 'C++' started by Aarti, Jul 27, 2007.

  1. Aarti

    Aarti Guest


    Can some one please explain why the output of this program is 15

    #include <iostream>

    using namespace std;

    class A
    int i;

    class B: public A
    int j;

    int f(A* p, int count)
    int total = 0;
    for(int i=0; i<count; ++i)


    int main()
    B b[11];
    cout << f(b,10);
    return 0;

    Aarti, Jul 27, 2007
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  2. You invoked undefined behaviour by treating an array of Bs as array of As.

    See Item 3 "Never treat arrays polymorphically" in Scott Meyers's "More
    Effective C++"
    Thomas J. Gritzan, Jul 27, 2007
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  3. Aarti

    Daniel T. Guest

    Note, sizeof(B) > sizeof(A). For the sake of this explanation, let's
    assume that sizeof(B) is 8 and sizeof(A) is 4.
    The increment above moves 'p' by sizeof(A) bytes. i.e., 4 bytes, but
    since the object pointed to by 'p' is really a B, 'p' now points to the
    middle of a B object.
    In the above line, you create an array of B. If sizeof(B) is 8, and b[0]
    is at address location 0x0, then b[1] is at address location 0x8.
    Daniel T., Jul 27, 2007
  4. Aarti

    joe Guest

    You have exercised a big no-no. When you increment a pointer, what
    you are effectively doing is adding the sizeof the pointee type to the
    value of the pointer. This allows you to visit the next element of an
    array automatically. What you have done here is passed an array of B-
    s (which are larger than A-s) into your function and trying to access
    it as though it were A-s. So, what happens is that you take your
    pointer and add sizeof A to it. This means that your pointer now
    points into the middle of your B object somewhere. It just so happens
    that since this class is simple and nothing faults, but you are now
    treating the middle of the B as another A. This is bad. It turns out
    that in your example you effectivly access b[0].i + b[0].j + b[1].i +
    b[1].j .... and so forth. This gives 5 * 1 + 5 * 2 = 15. Normally, I
    might expect a fault if things were more complicated than simple ints
    and it is definitely undefined behavior.

    joe, Jul 27, 2007
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