Defining a function argument as a pointer

Use the ‘&’ character in frnt of the function argument definition.

That’s it, there’s no other change needed, nothing is prepended to the variable when used or the function argument when called.

function MyFunction (&$MyVariable)
{
  $MyVariable = "abc" . $MyVariable . "hij";
}

$Name = def;

$Name = MyFunction($Name);
echo $Name;    //Will print "abcdefhij"

Array references

Works just the same as for variables.

Feel free to comment if you can add help to this page or point out issues and solutions you have found. I do not provide support on this site, if you need help with a problem head over to stack overflow.

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