About bound of array

Question

I find a piece of C++ code in textbook and have some questions.:

int ia[10]; // an integer array with 10 elements
int *ptr = ia; //the address of the first element in array. 
int *end = &ia[10]; //
while ( ptr != end ){
    std::cout<<*(ptr++)<<" ";
}
std::cout<<std::endl;

Since, the boundary of this array is from 0 to 9, which means that the ia[10] is out of boundary. Why it is allowed in C and C++?

Are you sure that is the code form text book b/c `ia[10]` is undefined behavior although pointing to one past the end is ok. — Shafik Yaghmour, Dec 16 '13 at 05:10
@chris: Seems to be valid C++, see the question Mark found for more information. — Ben Voigt, Dec 16 '13 at 05:14
@BenVoigt, I just did. Interesting, although it seems rather debatable judging by the...erm...debates :p — chris, Dec 16 '13 at 05:16
@allen.lao you originally had `ia[10]` but the corrected code is now valid. — Shafik Yaghmour, Dec 16 '13 at 05:16

Gavin · Answer 1 · 2013-12-16T05:32:06.913

1

end is pointing to the address immediately after the 10 elements of ia. The while loop is then looking to see when ptr, which is a pointer stepping along the array, reaches the address immediately after the array ia and thus the loop terminates. In this way there is no attempt to output any values beyond the initial ia array.

The key is that C uses references to memory locations.

Chapter 2 of this will give more info: http://pdos.csail.mit.edu/6.828/2012/readings/pointers.pdf

edited Dec 16 '13 at 05:32

answered Dec 16 '13 at 05:14

Gavin

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No problems. Check my edit for a link to more info. – Gavin Dec 16 '13 at 05:32

About bound of array

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