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Asked: 2026-05-11T14:37:14+00:00

So, as I learned from Michael Burr ‘s comments to this answer , the

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So, as I learned from Michael Burr‘s comments to this answer, the C standard doesn’t support integer subtraction from pointers past the first element in an array (which I suppose includes any allocated memory).

From section 6.5.6 of the combined C99 + TC1 + TC2 (pdf):

If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow; otherwise, the behavior is undefined.

I love pointer arithmetic, but this has never been something I’ve worried about before. I’ve always assumed that given:

 int a[1];  int * b = a - 3;  int * c = b + 3;

That c == a.

So while I believe I’ve done that sort of thing before, and not gotten bitten, it must have been due to the kindness of the various compilers I’ve worked with – that they’ve gone above and beyond what the standards require to make pointer arithmetic work the way I thought it did.

So my question is, how common is that? Are there commonly used compilers that don’t do that kindness for me? Is proper pointer arithmetic beyond the bounds of an array a defacto standard?

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  1. This is not ‘implementation defined’ by the Standard, this is ‘undefined’ by the Standard. Which means that you can’t count on a compiler supporting it, you can’t say, ‘well, this code is safe on compiler X’. By invoking undefined behavior, your program is undefined.

    The practical answer isn’t ‘how (where, when, on what compiler) can I get away with this’; the practical answer is ‘don’t do this’.

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