using count in if statement and comparing with its returned value doesn't work

Question

if(count($arr1) === count($arr2) === 26)

In the above line it is throwing error

syntax error, unexpected '==='

Why can't i compare the value returned by count() with 26

That's just invalid PHP syntax. You'll have to split up the comparisons and use `&&` — ಠ_ಠ, Jun 18 '13 at 12:30
Well, I think this should have worked. At first, `count($arr2) === 26` should be evaluated to boolean `true` or `false`. Then `count($arr1)` should be (strictly) compared to boolean, giving `false` in any case. So why does this produce error? (I don't mean this makes something different than OP wants, but it should be valid -- I vote up). — Voitcus, Jun 18 '13 at 12:33
@hek2mgl it's not low quality, many guys stucked with this http://stackoverflow.com/questions/1075534/cant-use-method-return-value-in-write-context?rq=1 — Mark Evans, Jun 18 '13 at 12:38
It's **really** low quality and a duplicate as you have shown. Sorry, please don't take it personal. Just try to fix your errors for yourself. If it not works try harder, investigate! but don't ask such questions on SO — hek2mgl, Jun 18 '13 at 12:49
@MarkEvans Yes I have, but no syntax error would be if second comparison is in brackets: `if($a == ($b == $c))` has no syntax error (doesn't matter if use `==` or `===`), but in `if($a == $b == $c)` there is one. So the accepted answer is not 100% correct. — Voitcus, Jun 18 '13 at 12:55
@MarkEvans I don't wanted to be offensive or something like that. I just wanted to say, if you want to get a programmer you'll have to learn to fix such problems for yourself. (Again, not offensive): Use the [manual](http://php.net/docs.php) it is **really** helpful. — hek2mgl, Jun 18 '13 at 12:56
@hek2mgl I think I really didn't made any efforts to investigate the problem and made some silly mistakes. — Mark Evans, Jun 18 '13 at 13:00
@MarkEvans Ok. :) However it is not so important anyway. it's just a question on SO (one out of 5 million :) .. You'll learn to fix such problems for yourself, I'm sure.. cu — hek2mgl, Jun 18 '13 at 13:03

score 4 · Answer 1 · edited Jun 18 '13 at 12:32

4

You can but you have to do it separately:

if(count($arr1) === 26 && count($arr2) === 26)

edited Jun 18 '13 at 12:32

Bathsheba

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answered Jun 18 '13 at 12:30

Tobias Golbs

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score 1 · Accepted Answer · answered Jun 18 '13 at 12:31

1

If count($arr2) is 26

if(count($arr1) === count($arr2) === 26)

evaluates to

if(count($arr1) === True)

which would then fail.

You need to perform two logic checks and check they both evaluate to true

answered Jun 18 '13 at 12:31

davbryn

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does it checks it backwards – Mark Evans Jun 18 '13 at 12:37
Yes, from right to left – davbryn Jun 18 '13 at 12:39
I've written the same in [my comment](http://stackoverflow.com/questions/17168889/using-count-in-if-statement-and-comparing-with-its-returned-value-doesnt-work#comment24857263_17168889), but there is a syntax error here. – Voitcus Jun 18 '13 at 12:44

using count in if statement and comparing with its returned value doesn't work

2 Answers2