How to check a double's bit pattern is 0x0 in a C++11 constexpr?

Question

I want to check that a given double/float variable has the actual bit pattern 0x0. Don't ask why, it's used in a function in Qt (qIsNull()) that I'd like to be constexpr.

The original code used a union:

union { double d; int64_t i; } u;
u.d = d;
return u.i == 0;

This doesn't work as a constexpr of course.

The next try was with reinterpret_cast:

return *reinterpret_cast<int64_t*>(&d) == 0;

But while that works as a constexpr in GCC 4.7, it fails (rightfully, b/c of pointer manipulation) in Clang 3.1.

The final idea was to go Alexandrescuesque and do this:

template <typename T1, typename T2>
union Converter {
    T1 t1;
    T2 t2;
    explicit constexpr Converter( T1 t1 ) : t1(t1) {}
    constexpr operator T2() const { return t2; }
};

// in qIsNull():
return Converter<double,int64_t>(d);

But that's not clever enough for Clang, either:

note: read of member 't2' of union with active member 't1' is not allowed in a constant expression
constexpr operator T2() const { return t2; }
                                       ^

Does anyone else have a good idea?

Esqueda

I guess there are other bit patterns that also represent a floating point 0, that you don't want to find? – Overstuff

There are exactly two bit patterns that represent 0. They are 000...000 and 100...000. The first bit is the sign bit. The second bit pattern is sometimes referred to as "negative zero". – Rep

Perhaps return d == 0 && 1/d > 0;? (en.wikipedia.org/wiki/Signed_zero#Comparisons) – Pressman

Don't ask why Don't tell me what to do – Lahdidah

@Lightness: don't tell Marc what not to do ;-p – Epizootic

reinterpret_cast is prohibited to appear in constant expressions. But what if you attempt static_cast<const int64_t*>(static_cast<const void*>(&d)) instead? – Strontianite

@LucDanton: I don't think it's reinterpret_cast vs. static_cast that's the issue, at least Clang rejects both. I guess it doesn't like taking the address. After all, how could it compute that at compile-time? – Esqueda

@jdv-JandeVaan: cf. the original code. It tests for a literal 0x0 bitpattern, not FP-equals-zero. I do wonder why they're so keen on skipping -0, but as a first approximation, I want to maintain behavioural compatibility while turning that thing constexpr... – Esqueda

@MarcMutz-mmutz Addressof is fine as a constant expression. It even appears in the (non-normative) examples: constexpr const int* addr(const int& ir) { return &ir; } // OK. reinterpret_cast, on the other hand, is explicitly forbidden. So obviously that's going to be a roadblock. – Strontianite

Answer 1

I want to check that a given double/float variable has the actual bit pattern 0x0

But if it's constexpr then it's not checking any variable, it's checking the value that this variable is statically determined to hold. That's why you aren't supposed to pull pointer and union tricks, "officially" there isn't any memory to point at.

If you can persuade your implementation to do non-trapping IEEE division-by-zero, then you could do something like:

return (d == 0) && (1 / d > 0)

Only +/-0 are equal to 0. 1/-0 is -Inf, which isn't greater than 0. 1/+0 is +Inf, which is. But I don't know how to make that non-trapping arithmetic happen.

Epizootic

Isn't the implementation usually non-trapping by default? – Brecciate

@Christian: on GCC with vanilla command-line options I get a floating point error if I divide by zero. – Epizootic

Aha, Ok. By the way, if I read correctly you can actually use feholdecxept from <cfenv> (C++11/C99) to enter non-trapping mode. And since he uses C++11, this might be a nice addition to your answer (but don't forget #pragma STDC FENV_ACCESS ON in this case). – Brecciate

Ah forget my comment, I forgot were talking about compile-time here. – Brecciate

Division by zero renders an expression non-constant in C++11. – Relate

Answer 2

It seems both clang++ 3.0 and g++ 4.7 (but not 4.6) treats std::signbit as constexpr.

return x == 0 && std::signbit(x) == 0;

Relay

Clang 3.0 doesn't implement constexpr? Anyway [c.math]/11 of C++11 clearly states that std::signbit isn't constexpr :( – Esqueda

Answer 3

It is not possible to look at the underlying bit pattern of a double from within a constant expression. There was a defect in the C++11 standard which allowed such inspection by casting via a void*, but that was addressed by C++ core issue 1312.

As "proof", clang's constexpr implementation (which is considered to be complete) has no mechanism for extracting the representation of a constant double value (other than via non-standard vector operations, and even then there is currently no way to inspect the result).

As others have suggested, if you know you will be targeting a platform which uses IEEE-754 floating point, 0x0 corresponds to the value positive zero. I believe the only way to detect this, that works inside a constant expression in both clang and g++, is to use __builtin_copysign:

constexpr bool isPosZero(double d) {
  return d == 0.0 && __builtin_copysign(1.0, d) == 1.0;
}

Relate

Answer 4

In c++20 one can use constexpr bit_cast, and so the check for the unsigned zero can be done like this:

std::bit_cast<uint64_t, double>(d) == 0

Stereochemistry