Hi,

just a guess: At compile time the compiler knows that the expression in the if statement will always evaluate to a constant 2, the if clause is probably meaningless, you get a warning. In the case of if ($foo = $bar) it may be intended, but anyway: The compiler can't evaluate the if clause to a constant as $bar can be true or false, so the conditional if may make sense.

Best regards
McA

In reply to Re: No warning when assiging to a variable by McA
in thread No warning when assiging to a variable by mikeraz

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