Result not displayed when comparison operation returns false

Question

I am trying to display the result of the comparison operation ($a > $b) which does not display anything for the below code, whereas the operation ($a < $b) displays the result 1.

I am wondering why the first operation does not return 0 as it is false?

<?php
    $a = 1;
    $b = 5;
    print ($a > $b);
    echo "Output";
    print ($a < $b);
?>

score 3 · Accepted Answer · answered Apr 12 '13 at 01:11

3

Unfortunately that's an extremely common misconception. 0 "is" not false. false is false, a boolean. 0 is 0, a number. 0 loosely equals false in a non-type safe comparison.

When you echo or print false, it is being cast to a string. false cast to a string is "", an empty string. true cast to a string is "1", the string "1".

Read http://php.net/manual/en/language.types.string.php#language.types.string.casting.

answered Apr 12 '13 at 01:11

deceze

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PHP has no objective standard for truth - it plays hard and fast with the truth. Truth is relative, according to PHP. "True and false are whatever I desire." – Patashu Apr 12 '13 at 01:13
@Patashu That's bullcr*p. PHP has perfectly defined [truth tables](http://www.php.net/manual/en/types.comparisons.php), you just have to learn them. PHP is not random. It boils down to the following things being *falsey*: `false`, `0`, `'0'`, `''`, `null`, `array()`. Everything else is *truthy*. – deceze Apr 12 '13 at 01:14
@Patashu That comment also has nothing to do with the topic at hand: *type juggling*. – deceze Apr 12 '13 at 01:22
You can memorize all of the weird comparison rules in PHP, but it's so easy to get wrong. For example http://phpsadness.com/sad/47 has some of my favourite gotchas, because 'string that contains a number' can suddenly become true when you don't expect it - `"1e1"` being a number for example... – Patashu Apr 12 '13 at 01:31
@Patashu But again, that's [perfectly defined behavior](http://www.php.net/manual/en/language.operators.comparison.php#language.operators.comparison.types). And again, you need to learn your language, especially if it has a lot of such automagic going on. It's not like you can't do anything about it and are at the mercy of PHP; you just have to know what your operators do and what the rules are. Every language has its type comparison gotchas, complaining about it doesn't make you a good programmer. – deceze Apr 12 '13 at 01:37

score 0 · Answer 2 · answered Apr 12 '13 at 01:11

0

Integer comparison ($a < $b) has return type boolean, which can be true or false.
print works in such way that it prints "1" for true and nothing for false.

answered Apr 12 '13 at 01:11

Martinsos

1,663
15
31

score 0 · Answer 3 · answered Apr 12 '13 at 01:13

0

It is due to the way in which PHP handles false values.

echo ($a > $b) will also output nothing.

To have your statement output 0 for a false condition you need to cast to an integer:

print (int)($a > $b);

answered Apr 12 '13 at 01:13

Ken Herbert

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Result not displayed when comparison operation returns false

3 Answers3