PHP Yield keyword - loop iteration

Question

I am using PHP generator and can't explain this behavior. This is the code I tried

<!-- language: PHP -->
<?php
function myfun($num1, $num2, $ctr = 1) {
    for ($i = $num1; $i <= $num2; $i =+ $ctr) {
        yield $i;
    }
}
echo 'Odd numbers: ';
foreach(myfun(1, 7, 2) as $num) {echo $num;};
?>

Can someone explain me this behavior using PHP yield, entering a infinite loop?

result: Odd numbers: 122222222222222222222222222222222...............

Note: $i += $ctr works as expected

result: Odd numbers: 1357

$i =+ $ctr is the same as $i = $ctr. + some number is essentially a no-op. — georg
– georg, Commented Oct 13, 2014 at 14:09

naneau · Accepted Answer · 2014-10-13 14:10:58Z

1

The problem lies in the =+ operation, you probably meant to type +=, which would do the trick:

<?php
function myfun($num1, $num2, $ctr = 1) {
    for ($i = $num1; $i <= $num2; $i += $ctr) {
        yield $i;
    }
}
echo 'Odd numbers: ';
foreach(myfun(1, 7, 2) as $num) {echo $num;};

Result: Odd numbers: 1357

answered Oct 13, 2014 at 14:10

naneau

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4 Comments

Jean-Christophe Verhulst Over a year ago

I agree with this, but why 1222222, and not 1111111??

naneau Over a year ago

you assign $i as $num1 in the first iteration, and you call it with 1 as a value for $num1

Jean-Christophe Verhulst Over a year ago

1 is the default value in case of not provided; Remove =1 (so function myfun($num1, $num2, $ctr)) in the loop definition, same behavior!!!

naneau Over a year ago

In the first iteration $i is equal to $num1, that's the first part of the for(). In the second iteration (and all others), it will be set to $ctr (which is 2). So you get one loop of $i being 1, and all the next ones where it will be 2. The loop doesn't finish, because $i <= $num2 is never true.

Jim · Accepted Answer · 2014-10-13 14:18:33Z

1

$i =+ $ctr

=+ is not an operator. This will essentially do $i = $ctr.

The first time the loop occurs $i is set to $ctr, in this case this is 2. After this it is continually set to 2 and never goes higher. Hence the infinite loop. Use += instead.

answered Oct 13, 2014 at 14:18

Jim

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1 Comment

Agnius Vasiliauskas Over a year ago

@Jean-ChristopheVerhulst Don't blame the parser. Nothing to complain for here - infinite loop is a valid syntactic construct. And here + symbol is interpreted as unary plus operator, i.e. $x = 10 and $x = +10 are semantically equivalent. Parser can't know that you instead of unary plus operator wanted to write compound assignment operator. So ...

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