Thread: typedef modifiers

  1. #1
    Registered User SKeane's Avatar
    Join Date
    Sep 2006

    typedef modifiers

    Simple question really, why is the following not legitimate syntax?

    typedef long long foo_t;
    foo_t f;
    unsigned foo_t uf;
    And yes, I do realise I can do ...

    typedef long long foo_t;
    typedef unsigned long long ufoo_t;
    foo_t f;
    ufoo_t uf;
    I was just curious why the first syntax isn't allowed.
    Last edited by SKeane; 09-26-2006 at 03:36 AM.

  2. #2
    Registered User
    Join Date
    Jun 2005
    Because the typedef creates foo_t as a synonym for a type named "long long", which is more precisely described as "long long signed". A basic rule is that no declaration can have more than one of the set {long, short} or of the set {signed, unsigned} as type modifiers (eg it is not possible to declare something of type "long short" or "short short").

    "long long" is actually an allowed exception to that rule introduced by more recent standards ("long long" is treated as a single type modifier rather than a pair) but that is essentially how it works --- declaring something as a "unsigned signed long long int" doesn't make sense.

    In terms of the language grammar, the "typedef" keyword is treated as another storage modifier, and the original rule was that it is not possible to add other storage modifiers when using a typedef'd type. However, there are a few exceptions now built into the grammar to accomodate the "long long" type, so my description above is functionally correct, if not necessarily how a compiler does it.
    Last edited by grumpy; 09-26-2006 at 05:24 AM.

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