Question

So my code in the past needed a variable to run through 2 functions and then return the value as such.

 function 1($variable) {
    check($variable);
    return $variable // either a -1, -2 or true;
 }

 // pass the return to next function

function 2($variable) {
    check($variable);
    return $variable // either a -1, -2 or true;
}

On the next check it returns a message to the user as such:

  if($variable == -1) // display message
  if($variable == -2) // display message
  if($variable == true) // display message

Now, per requirement of work the variable must go through a 3rd function check still returning a -1, -2 or true and then go onto the final if statements for display.
Now this is where it gets odd. If I keep it at 2 functions the if statements work, however if I run it through the 3rd check function I need to format my if’s like this in order to correctly check the return:

if($variable === -1) // display message
if($variable === -2) // display message
if($variable === true) // display message

Notice I have to add the 3rd ‘=’ symbol. I just can’t figure out why this is happening. Is this normal by some PHP law I don’t know about or is this a bug?

Answer 1

This is not odd behavior, it’s very natural for PHP.

Following expression:

if ($variable == true) {
}

means that PHP will cast left operand to less pretensive type(in this case BOOLEAN) and do comparison after this. Which obviously will result in TRUE if $variable value is not 0 or FALSE or NULL or ”

In second case i.e. === there is strict check value and type of both operands are compared.