Why is this code an infinite loop?

Question

Before completing this code, I just tested it by mistake and realized that it will not stop:

$var = "any"; 
for ($i=1; $i < 2; $i++){ 
    $var.$i = "any"; 
}

Why does this produce an infinite loop? And why doesn't PHP produce an error?

score 5 · Accepted Answer · edited Jul 25 '13 at 13:02

5

I did a simple test :

echo $i;
 $var.$i = "any";
var_dump($var);

Result :

1string(3) "any"
anzstring(3) "any"

So $i get transformed to "anz" and doesn't pass the validation to get out of the loop.

$var.$i = "any"; is not really correct, i don't know what you are trying to do, but if you want to fill and array you should do something more like :

$var = array();
for ($i=1; $i < 2; $i++){ 
 $var[] = "any";
}

If you want to change your string letter by letter :

$var = "any";
    for ($i=1; $i < 2; $i++){ 
     $var[$i] = "a"; // asign a new letter to the string at the $i position
    }

edited Jul 25 '13 at 13:02

answered Jul 25 '13 at 11:08

Morsok

126
1
3

thanks morsok, I knew these, I just wonder why PHP not produce any error, if this `$var.$i` not valid then why PHP executes? – Jul 25 '13 at 11:11
Strictly speaking it's valid, the behavior is simply not what you expect. Internally i can't tell you exactly what it does, but it seems to change $i to the string "any" with the last letter being added the value of $i making 'y' go 'z', and this is always < 2 hense it's goes to an infinite loop. – Morsok Jul 25 '13 at 11:14
@Skippy But I don't see any `z` here, ok, can you please explain what is `z`? – Jul 25 '13 at 13:04
1

ok any way, thanks for your attention, I hope he decide if its z or y – Jul 25 '13 at 13:08
1

It's a 'z', it didn't add space to the debug sorry about that, you should read the debug as : echo $i = "anz" var_dump($var) = .... – Morsok Jul 25 '13 at 13:19

score 2 · Answer 2 · answered Jul 25 '13 at 11:12

When you do the following $var.$i = 'any' you set the $i variable and the $var variable. So the the loop never stop running because var_dump($i < 1) returns true.

$var = 'var';
$i = 1;
$var.$i = 'var';

var_dump($i);

Returns string(3) "var".

This loop will never stop because $i is always reset to 'var', which is smaller than 1.

score 1 · Answer 3 · edited Jul 25 '13 at 17:30

1

This is incorrect $var.$i = "any"; because this expression is equivalent to:

 $var.($i = "any");

Which assigns $i to new value, therefore the condition of which the while loops checks will always be true.

edited Jul 25 '13 at 17:30

answered Jul 25 '13 at 11:03

Maciej A. Czyzewski

1,539
1
13
24

I knew that, what is incorrect? why makes the loop go for Infinity! – Jul 25 '13 at 11:03

score 0 · Answer 4 · answered Jul 26 '13 at 09:55

PHP5.4+.

You will get 'anz' result, after $i++, when $i == 'any'. $i == 'any', after assignment and that is, actually, what it should get. Trick is in "$i='any'" part of line. Even when "=" has lower precedence then ".", why do you think that it shouldn't put 'any' inside $i?

Try this instead:

$var = "any"; 
for ($i=1; $i < 2; $i++){ 
 $i.$var = "anything";
}

And your loop will work. And $var will get "anything" value. This doesn't look like a bug. Just unexpected behaviour for somebody.

Why is this code an infinite loop?

4 Answers4

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