Consider the code:
int arr[20]{};
int * ptr1=arr, * ptr2=&arr[1];
std::cout<<ptr1<<std::endl<<ptr2<<std::endl<<ptr2-ptr1;
Output:
0x7fff4003e0d0
0x7fff4003e0d4
1
Why it isn't 4 instead?
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dlpsankhla
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2Because it is the definition of pointers substraction? – MikeCAT Jun 08 '16 at 15:36
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Because type system. – Kerrek SB Jun 08 '16 at 15:37
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Isn't `arr[1]` 1 away from `arr[0]`(`arr`)? – NathanOliver Jun 08 '16 at 15:37
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@MikeCAT Then how can I get the result as 4 in any manner? – dlpsankhla Jun 08 '16 at 15:46
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By multiplying by 4? – MikeCAT Jun 08 '16 at 15:53
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@MikeCAT I mean if I do this : cout<<0x7fff4003e0d4 -0x7fff4003e0d0; assuming that there is no any array declaration, then why it is yielding 4 but here it is 1? – dlpsankhla Jun 08 '16 at 15:56
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Because `0x7fff4003e0d4 -0x7fff4003e0d0` is just a normal integer aritimetic, not pointer aritimetic. – MikeCAT Jun 08 '16 at 16:00
1 Answers
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If it didn't to that you'd just end up dividing the difference by sizeof(whatever) all the time. The number of elements is far more useful than the raw difference. When you need the latter, cast the two pointers to char* for the subtraction.
Pete Becker
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I mean if I do this : cout<<0x7fff4003e0d4 -0x7fff4003e0d0; assuming that there is no any array declaration, then why it is yielding 4 but here it is 1? – dlpsankhla Jun 08 '16 at 15:57