c++ pointer arithmetic

Question

int a[5];

cout << &a[1] << " " << &a[0] << endl;
cout << (&a[1] - &a[0]);

In the above code, why is &a[1] - &a[0] equal to 1 and not 4? Shouldn't there be 4 bytes between these addresses since we have an int array?

NPE · Accepted Answer · 2011-11-25 20:54:14Z

8

No, pointer difference is in elements, not in bytes.

answered Nov 25, 2011 at 20:54

NPE

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1 Comment

simple as that. @user974967: reading the book will be cheaper in the long run

Adrian Cornish · Accepted Answer · 2011-11-25 20:56:42Z

2

Pointers are incremented by the size of there type. Reason is because you want to point to the next item. So taking you example further.

int a[5];
int *ptr=&a[0];

// ptr is now pointing at first element.

ptr+3; // now its pointing at 3rd element.

answered Nov 25, 2011 at 20:56

Adrian Cornish

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sehe · Accepted Answer · 2011-11-25 20:56:37Z

1

To get it in bytes: (see it live https://ideone.com/CrL4z)

int a[5];

cout << (a+1) << " " << (a+0) << endl;
cout << (reinterpret_cast<char*>(a+1) - reinterpret_cast<char*>(a+0));

answered Nov 25, 2011 at 20:56

sehe

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