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Request #71758 can't initialize inline a pass-by-references variable parameter
Submitted: 2016-03-09 15:36 UTC Modified: -
Votes:1
Avg. Score:5.0 ± 0.0
Reproduced:1 of 1 (100.0%)
Same Version:1 (100.0%)
Same OS:1 (100.0%)
From: vittorio dot zamparella at gmail dot com Assigned:
Status: Open Package: Compile Warning
PHP Version: 7.0.4 OS:
Private report: No CVE-ID: None
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 [2016-03-09 15:36 UTC] vittorio dot zamparella at gmail dot com
Description:
------------
While I perfectly understand why I cannot pass by reference a literal, I whish that I could initialize a variable in the same place where I pass it as a parameter.
MyFunction( $var2 = "two" );

I think it is more readable, understandable and inline with the syntax of the language.
Maybe I'm missing some symple syntactic sugar to work around this (apart from initializing the variable the line before).

I understand that the result of assignment is an anonimous literal with containing the just assigned value (hence the warning), but I think that the compiler could be just that little bit smarter to allow for such a syntax, since the semantic is perfectly clear: assign the literal to the variable, then pass the variable (and not its value to the function.
Also the algorithm is quite simple: if the actual parameter is a literal and a variable is expected, if the literal is the product of an assignment pass the left part of the assignment to the function.
Actually I don't know if the parser can trace back where a value is coming from.

Thank you for your comments and explanantions.

Test script:
---------------
<?php
function MyFunction (&$parameter) { 
    echo $parameter."\n";
}
$var1 = 'one';
MyFunction( $var1 );
MyFunction( $var2 = "two" );

Expected result:
----------------
one
two

Actual result:
--------------
one

Notice: Only variables should be passed by reference in /in/iJAlU on line 7
two

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