RE: [PHP-DB] weird comparsion

On 03 May 2007 12:30, OKi98 wrote: > I know about identity operator (===) but with == operator 0 is false > and "foo" is true No, that's not correct. , try this: > > $foo=0; > $bar="bar"; > if ($foo) echo("$foo is true, "); > else echo("$foo is false, "); > if ($bar) echo("$bar is true, "); > else echo("$bar is false, "); > if ($foo==$bar) echo("$foo==$bar"); > > returns "0 is false, bar is true, 0==$bar" That's because you've got loads of implicit type conversions going on there, so you're not comparing like with like. For if ($foo) ... and if ($bar) ...: within the context of the if(), both $foo and $bar are implicitly converted to Boolean: - (bool)0 is FALSE - (bool)"any non-empty() string" is TRUE On the other hand, for if ($foo==$bar) ...: in the context of the == comparison, $bar is converted to a number, and any string not beginning with a numeric character converts to numeric zero -- so you get a comparison of zero with zero, which is, of course, TRUE! Or, in other words, (bool)$foo!==(bool)$bar, BUT (int)$foo===(int)$bar It's exactly when you *don't* want this kind of automatic type-conversion shenanigans going on that you should use the === operator to make your intent entirely clear -- otherwise you have to be extremely aware of the context in which you are evaluating your variables in order to avoid hidden surprises like this. Cheers! Mike --------------------------------------------------------------------- Mike Ford, Electronic Information Services Adviser, JG125, The Headingley Library, James Graham Building, Leeds Metropolitan University, Headingley Campus, LEEDS, LS6 3QS, United Kingdom Email: m.ford@leedsmet.ac.uk Tel: +44 113 812 4730 Fax: +44 113 812 3211 To view the terms under which this email is distributed, please go to http://disclaimer.leedsmet.ac.uk/email.htm