A problem about in_array
arrays, php
Solution
It means you have to set the third parameter to `true` when you want the comparison to not only compare values, but also types.
Else, there is type conversions, while doing the comparisons -- see String conversion to numbers, for instance.
As a matter of fact, `in_array` without and with strict is just the same difference as you'll have between `==` and `===` -- see Comparison Operators.
This conversion, most of the time, works OK... But not in the case you're trying to compare 0 with a string that starts with a letter : the string gets converted to a numeric, which has 0 as value.
Problem
I have got a strange problem about `in_array` recently which I cannot understand. e.g. ``` $a = array('a','b','c'); $b = array(1,2,3); if (in_array(0,$a)) { echo "a bingo!\n"; } else { echo "a miss!\n"; } if (in_array(0,$b)) { echo "b bingo!\n"; } else { echo "b miss!\n"; } ``` I ran it on my lamp,and got ``` a bingo! b miss! ``` I read the manual and set the third parameter `$strict` as `true`,then it worked as expected.But does that mean I always need to set the strict parameter as true when using `in_array`?Suggestions would be appreciated. Regards