pointer arithmetic(pointing to arrays)

Question

i've a small doubt in pointers, please help me out..

void main()
{
    int x[10],*px=x,*py;
    int i;
    py = &x[5], i = py - (px);

    cout << "\nThe value of px=x is:" << (int)px << "\n";
    cout << "x[0]\t" << (int)x << "\n";
    cout << "x[5]\t" << (int)&x[5] << "\n";
    cout << "\nThe value of i=py-px is\n";
    cout << i;
}

in the above program, you get the value of 'i' as the difference of the integer equivalent of the array(memory) divided by two(10/2=5).Why is it not just the difference ie, 10??

thanks in advance!!

It sounds like you expect sizeof(int)=2. Can you explain why you would expect the result to be 10? — user1157123
– user1157123, Commented Feb 19, 2013 at 15:57
well, suppose you get the integer eqv of memory of x[5] as '-24' and x[0] as '-34'..then shouldn't px-py result in -24+34=10?? — vpp
– vpp, Commented Feb 19, 2013 at 16:05

twalberg · Accepted Answer · 2013-02-19 16:19:31Z

1

If you are trying to get the difference between two of the array elements using your pointers, you need to dereference the pointers:

i = *py - *px;

The way you have it written, you are calculating the difference between the two addresses, which should be 5, unless you cast your pointers to void *, in which case it would be 5 * sizeof(int) (not sure if you're on 32-bit or 64-bit system - the answer would be different).

Oh, and you're not initializing x[] anyway, so your results might be a bit ... random...

answered Feb 19, 2013 at 16:19

twalberg

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