Clarification on Gcc's strict aliasing rules

Francis Moreau francis.moro@gmail.com
Fri Nov 12 16:01:00 GMT 2010


"Segher Boessenkool" <segher@kernel.crashing.org> writes:

>>>> 	  int f() {
>>>> 	    union a_union t;
>>>> 	    int* ip;
>
>>>> 	    t.d = 3.0;
>>>> 	    ip = &t.i;
>>>> 	    return *ip;
>>>> 	  }
>>>>
>>>> could you tell me what the effective type of 't.i' object ?
>>>
>>> int, if you can say that object exists at all: it does not have a stored
>>> value.  The stored value of t is a double with value 3.0 .  You can
>>> take its address and access it via that as "double" (or "char"), or you
>>> can access it as the union it is.  You can not access it as "int".
>>
>> BTW, does your reasoning rely on the C standard ?
>
> Of course it does.  Perhaps I don't understand what you're asking here.
>

Ok, if I understand correctly you previous post, you say that after the
following expression statement:

    t.d = 3.0;

the stored value of t is a double and I agree.

You also said and I still agree that:

    - you can take its adress and access it via that as double (or
      "char"):

        double *d = &t.d;

    - you can access it as the union it is (I'm not sure if that what
      you meant):

        double d = t.d;  

But you finally said

    - you can not access it as int:

        that object (t.i) does not have a stored value therefore it
        doesn't exist.

This is what I understood from what you said, please correct me if I'm wrong.

However doing:

     int i = t.i;

is defined in C (as long as there's no trap representation) even if 't.i'
object has no stored value.

So I assume that 't.i' always exists, and doing:

     int *i = &t.i;

6.5p6 said that the effective type of 't.i' is 'int' whatever the value
stored in 't'.

-- 
Francis



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