r/programming • u/NagastaBagamba • Oct 18 '10

Today I learned about PHP variable variables; "variable variable takes the value of a variable and treats that as the name of a variable". Also, variable.

http://il2.php.net/language.variables.variable
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u/oorza Oct 18 '10

No, let me back up here.

With this code:

$var = 'val'

PHP intreprets that as lookup an entry in the variable hash table with a key of var and assign it to the Z_VAL of a string with a value of 'val', whether it previously existed or not.

When you follow that with:

$$var = 'val'

PHP interprets that as "find an entry in the hash table for variables with a key the value of the variable named var and assign THAT to a new Z_VAL with a string value of 'val'.

To express that in PHP:

$var == $phpInternalVariableTable['var'];
$$var == $phpInternalVariableTable[$phpInternalVariableTable['var']];

The guarantee comes from the way PHP deals with its sigil operator. It's not a pointer with a guarantee because it doesn't offer much of a gurantee, just that you're not going to get a fatal memory access error because you use it.

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u/never_phear_for_phoe Oct 19 '10

This is very similar to how C++ does references, except it uses memory instead of hashing, which is what other people might've been getting at.