is a[i++] = 3 undefined?

Question

I'm a novice programmer (been learning C on a CS course since September) I've been reading the following two questions in order to try and get my head around sequence points and the undefined behaviours which relate to them.

Why are these constructs (using ++) undefined behavior?

Why is a = i + i++ undefined and not unspecified behaviour

I now understand that

a[i] = i++;

results in an undefined behaviour. I was wondering if

a[i++] = 4;

falls into the same category but I can't find anyone discussing this specific example anywhere.

[No. It isn't UB](http://stackoverflow.com/a/32798674/3049655) — Spikatrix, Jan 24 '16 at 09:49
@CoolGuy I don't think this is a duplicate; if you read the question you suggested, the question body doesn't seem to relate to the title... — Ben Wainwright, Jan 24 '16 at 09:59
@BenWainwright Yes. But, I remember a similar question being asked and gained a lot of votes but now, can't seem to find it... — Spikatrix, Jan 24 '16 at 10:30
@CoolGuy ok, but for now could you remove that duplicate link as it is misleading :) — Ben Wainwright, Jan 24 '16 at 10:31

score 6 · Answer 1 · answered Jan 24 '16 at 09:49

6

No, the latter is not undefined behavior because you're not modifying and accessing i more than once between two sequence points.

answered Jan 24 '16 at 09:49

Filipe Gonçalves

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score 2 · Accepted Answer · answered Jan 24 '16 at 09:51

That's fine.

The problem with a[i] = i++; is that the modification in i++ is separate from the read in i in a[i]. The modification could take place before, during or after that read.

Since a[i++] = 4; doesn't read i independently from its modification, you don't have that problem.

is a[i++] = 3 undefined?

2 Answers2