PHP IF-ELSE 语句

Question

Code:

<?php
$a = 200;
$b = 300;
if ($a > $b + $b != 3)
    print "Correct";

else
print "Incorrect";
?>

output is: Correct

Can someone help me understand why the output became "Correct"?

Answer 1

To understand what is happening here, you need to look at the list ofoperator precendence to see what is being evaluated first. It's not left to right. The order of the operators in your if statement are as follows:

1. + - ++ -- ~ (int) (float) (string) (array) (object) (bool) @ - arithmetic (unary + and -), increment/decrement, bitwise, type casting and error control
2. < <= > >= - associative comparison
3. == != === !== <> <=> - non associative comparison

So in essence, your if statement breaks down to this:

(($a > ($b + $b)) != 3)

With your values becomes

((200 > (300 + 300)) != 3)
((200 > 600) != 3)
(false != 3)

So of course, false is not 3, and makes your if statement correct. If you want to evaluation 200 is greater than 300 AND 300 is not 3 , then you need the logical AND operator, or && , which would be

($a > $b && $b != 3)

which would print Incorrect