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Bug #289 PHP does not evaluate the () (Function call) operator correctly
Submitted: 1998-04-20 12:06 UTC Modified: 1998-04-24 10:23 UTC
From: kk at shonline dot de Assigned:
Status: Closed Package: Performance problem
PHP Version: 3.0 Release Candidate 4 OS: Solaris 2.5.1
Private report: No CVE-ID: None
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 [1998-04-20 12:06 UTC] kk at shonline dot de
Function calls with a variable name work as exspected when
called by a simple expression:

function name_lang() {
        printf("I am a function\n");
$x = "name";
$y = "_lang";
$z = $x.$y;
Content-type: text/html
I am a function

But PHP fails to parse the () function call operator in the
following environment:

<b>Parse error</b>:  parse error in <b>-</b> on line <b>11</b><br>


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 [1998-04-24 10:23 UTC] zeev
This syntax is not supported.  Only strings
and variables can be used as function names (not any
other expressions)

You can use
$foo = $x.$y;

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