Oh and that presumably applies equally for std::byte?

Which then makes me wonder: what is then the whole point is of having these types that can alias other types, if you can't use them for actually accessing any objects?

Cheers,
Timur

> On 21 Aug 2019, at 21:44, Timur Doumler <cpp_at_[hidden]> wrote:
>
> So you're saying that, even without any pointer arithmetic, just this code:
>
> int x = 100000;
> std::cout << *reinterpret_cast<char*>(&x);
>
> has undefined behaviour?
>
> If that's the case then this is even more insane than I thought. Please clarify whether this is really what you're saying here!
>
> Cheers,
> Timur
>
>> On 21 Aug 2019, at 21:33, sdkrystian via Std-Proposals <std-proposals_at_[hidden]> wrote:
>>
>> Ah, ignore my example then.
>>
>>
>>
>> Sent from my Samsung Galaxy smartphone.
>>
>> -------- Original message --------
>> From: "language.lawyer--- via Std-Proposals" <std-proposals_at_[hidden]>
>> Date: 8/21/19 15:31 (GMT-05:00)
>> To: std-proposals_at_[hidden]
>> Cc: language.lawyer_at_[hidden]
>> Subject: Re: [std-proposals] Allowing access to object representations
>>
>> On 21/08/2019 22:11, Timur Doumler wrote:
>> > This doesn't make sense to me.
>>
>> That's why some people think this should be fixed.
>>
>> > If I have an object of type int (or, say, a struct containing an int), then the value has 4 bytes. If I access that value by dereferencing a char pointer that aliases that object, all I can ever get from that char is a single byte. How can this be "the object"?
>>
>> The standard just says what result you get, it is an implementation's problem to answer the "how" question.
>> However, http://eel.is/c++draft/expr.pre#4 could help some implementations:
>> > If during the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined.
>>
>> If you have an int object and try to access its value through a char lvalue and the value stored in the int object is not in the range representable by the char type, the behavior is undefined.
>>
>> [expr.pre]/4 helps little-endian machine implementations, because they just can fetch the first byte of the int object and this would give the right value when there is no UB.
>> IDK how a big-endian maching implementation can achieve the standard-conforming behavior :-)
>>
>> > Cheers,
>> > Timur
>> >
>> >> On 21 Aug 2019, at 20:59, language.lawyer--- via Std-Proposals <std-proposals_at_[hidden]> wrote:
>> >>
>> >>> On 21/08/2019 21:55, Krystian Stasiowski via Std-Proposals wrote:
>> >>> Yes, you access the value of the object of the object, however, since its not specified what exactly you get
>> >>
>> >> It is specified: you get the value contained in the object.
>> >>
>> >>> http://eel.is/c++draft/conv.lval#3.4
>> >>
>> >> Indeed, it is specified here.
>> >> --
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>> >
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