Re: BOOST_IS_CONSTANT_EVALUATED_INT(integer-expression)(integer-expression)

From: Robert Ramey (ramey_at_[hidden])
Date: 2021-08-24 21:28:19

On 8/24/21 11:33 AM, Ivan Matek via Boost wrote:
> On Tue, Aug 24, 2021 at 6:01 PM Robert Ramey via Boost <
> boost_at_[hidden]> wrote:
> 
>> I guess if it doesn't work, it's not valid.  Certainly from what little
>> user documentation there is - cpp reference, there's no information on
>> this topic.  My user case looks like:
>>
>> template<typename T>
>> struct interval {
>>       T m_min;
>>       T m_ max;
>>       interval(const T & min, const T & max) :
>>           m_min(min),
>>           m_max(max)
>>       {}
>> }
>>
>> template<typename T>
>> constexpr interval<T> get_interval(const T & t) {
>>       return std::is_constant_evaluated()
>>           ? interval<T>(t, t); // we know the value of t at compile time
>> so the range is effectively [t, t]
>>           : interval<T>(std::numeric_limits<T>::min(),
>> std::numeric_limits<T>::min()); // otherwise we won't know the value
>> until runtime so it could be in the range [...]
>>       ;
>> }
>>
>> So I'm returning different values depending on whether we're compiling
>> or execution.  In fact, if we know the value of t at compile time, the
>> runtime code will never be executed.  So there would never be an
>> ambiguity, but of course the compiler can't see that so maybe that's why
>> the compiler is unhappy with this.
>>
> Still not 100% clear what bigger goal is, for me example code compiles
> <https://www.godbolt.org/z/sbKGh53ec> when I make constructor constexpr.

Hmmm - it wasn't clear to me that that each function has to be constexpr 
if the class is.

> I presume you want to use the returned interval to construct some types(for
> example something like safe_signed_range ) and that is where the problems
> may occur.
Correct.  Not that the type being returned is always the same.  It's 
just the values of the member variables which change.

> As far as I know this is impossible in C++, to have constexpr arguments,
> unless you use them as template arguments, e.g.  func<3>(), but if you have
> func(3) there is no way to get that 3 during compile time.

Given the problems the compiler has with this code, looks like you're 
correct.
> So even if this new boost macro is added that will not help you since even
> if compiler knows that 3 is 3 it will not let you to use it at compile time
> due to C++ language rules <https://www.godbolt.org/z/hWPTGvPss>.

Here's a much simpler example.  In this case, it doesn't look like it 
doesn't work on clang ever.

https://www.godbolt.org/z/c4s61fTvd

> So I am not an expert, but I think it is indeed not possible to do what you
> want.

It's not clear to my why not.  I'll dig into this some more.  I think I 
can achieve what I need without using a function.

Thanks to everyone for their interest and help.

Robert Ramey