PHP loose typing in while loop comparison - php

Question

Given the following code snippet:
$i= 11;
function get_num() {
global $i;
return (--$i >= 0) ? $i : false;
}
while($num = get_num()) {
echo "Number: $num\n";
}
Results in the following output:
Number: 10
Number: 9
Number: 8
Number: 7
Number: 6
Number: 5
Number: 4
Number: 3
Number: 2
Number: 1
However, I also want it to output Number: 0 - but the while loop evaluates 0 as being false, so the loop never gets to that point. How do I get the loop to terminate only on an explicit false?

Answer 1

while( ($num = get_num()) !== false ) {
extra = forces type check as well.

Answer 2

<?php
$i= 11;
function get_num() {
global $i;
return (--$i >= 0) ? $i : false;
}
while(($num = get_num())!==false) {
echo "Number: $num\n";
}
?>

Answer 3

You have to do a comparison that compares the types, and not only the values -- which means using the === or !== operators, instead of == or !=.
So, you could use something like :
while(($num = get_num()) !== false) {
echo "Number: $num\n";
}
With this, 0 will not be considered as the same thing as false.
As reference : Comparison Operators (quoting)
$a == $b : Equal : TRUE if
$a is equal to $b. $a === $b
: Identical : TRUE if $a is
equal to $b, and they are of the
same type.

Answer 4

I noticed that you are using global. Some devs scream about using global when not necessary. Also, you can leave out the parentheses if you write false!== before your $num declaration (no big deal, just a note).
Option 1
function get_num() {
static $i=11; // effectively only declare the value once
return (--$i >= 0) ? $i : false;
}
while(false!==$num=get_num()){
echo "Number: $num\n"; // $i is NOT available outside the scope of the function.
}
Option 2
function get_num($i) {
return (--$i >= 0) ? $i : false;
}
$num=11; // declare in global scope
while(false!==$num=get_num($num)){ // pass it to function as an argument
echo "Number: $num\n";
}