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Consider using a different name than the imported variable $fn, or did you forget to import by reference?
It seems like you are assigning to a variable which was imported through a use
statement which was not imported by reference.
For clarity, we suggest to use a different name or import by reference depending
on whether you would like to have the change visibile in outer-scope.
Change not visible in outer-scope
$x=1;$callable=function()use($x){$x=2;// Not visible in outer scope. If you would like this, how// about using a different variable name than $x?};$callable();var_dump($x);// integer(1)
Both the $myVar assignment in line 1 and the $higher assignment in line 2
are dead. The first because $myVar is never used and the second because
$higher is always overwritten for every possible time line.
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};
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}
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/**
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* Run tasks in series
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* @param \Closure[] $tasks
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* @param array $args
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* @return void
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*/
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private function waterfall (array $tasks, array $args)
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{
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$index = 0;
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$next = function () use (&$index, &$tasks, &$next, &$args) {
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if ($index == count($tasks)) {
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return;
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}
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$callback = $this->callOnce(function () use (&$next) {
It seems like you are assigning to a variable which was imported through a
use
statement which was not imported by reference.For clarity, we suggest to use a different name or import by reference depending on whether you would like to have the change visibile in outer-scope.
Change not visible in outer-scope
Change visible in outer-scope