Description:
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Any variable identifier can be used in a statement - even if it hasn't been referenced before - without producing an error.

Most of the time when debuggong php code is spent for searching typos in variable names.

I consider it a bug that php does not produce and display error if it encounters an unreferenced variable in a statement.

A solution could be a "strict" statement. That tells php to throw errors in that case to stay backward compatible.

An even better solution would be the need to 

Reproduce code:
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$nValue1 = 10;
$nValue2 = 20;
print($nVale1 + $nValue2);

Expected result:
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Expected:

30

No, it is:

20

Why, because of the typo in line 3. What I want to happen in that case is this:

Parse error: parse error in /somefile.php4 on line 3: unreferenced object "$nVale1"

Actual result:
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20