constexpr does not work/apply inside function call

Question

I have implemented a constexpr compile-time hash-function, which works fine (i.e. is evaluated at compile-time) if called as

constexpr auto hash = CompileTimeHash( "aha" );

but I need to use it in actual code as an argument to a function as in

foo( CompileTimeHash( "aha" ) ); // foo is NOT constexpr

For a specific reason, I cannot use the long version

constexpr auto hash = CompileTimeHash( "aha" );
foo( hash );

The compiler (VC++) will not compile-time hash in the short (first) case. Is there any way to achieve this?

EDIT: An example covering the 3 cases is now found here: https://godbolt.org/z/JGAyuE Only gcc gets it done in all 3 cases

If nothing else works, you can hack your compiler a bit by `foo([]{ constexpr auto hash = CompileTimeHash( "aha" ); return hash; }())` — lisyarus, Aug 24 '18 at 12:24
How do you *know* that the hash function is evaluated at run-time? — Some programmer dude, Aug 24 '18 at 12:31
@Someprogrammerdude "How do you know that the hash function is evaluated at run-time?" You can check by calling it within a static_assert — Petok Lorand, Aug 24 '18 at 12:38
Which version of VC++ are you using? Also,could we get a [mcve] to test against other compilers? — NathanOliver, Aug 24 '18 at 12:44
@Someprogrammerdude you [look at the disassembly](https://godbolt.org/z/7nmnW0) — n. m. could be an AI, Aug 24 '18 at 12:55
@n.m.: So it looks like a compiler bug / missed optimization opportunity, doesn't it? — einpoklum, Aug 24 '18 at 13:01
Wait, a breakpoint in debug or release builds? Debug is not relevant; release is untrustworthy. So again, how do you know? nm makes a strong case tho. — Yakk - Adam Nevraumont, Aug 24 '18 at 13:04
You should say more about the "specific reason" you "cannot use the long version"; just to know if that reason is an obstacle for a solution. Other question: what type return `CompileTimeHash()`? — max66, Aug 24 '18 at 13:12
@Yakk-AdamNevraumont: It is at least 100% safe to say, that if the breakpoint triggered (as it did), the code got executed at runtime. If it does NOT trigger, however, we cannot conclude that the code was only used at compile time. — old123987, Aug 27 '18 at 12:26
@old123987 No, breakpoints can trigger even if the code was optimized to the point where the stuff run is almost completely unrelated to the C++ code. Ie, the code in theory could be optimized to "load constant value" and yet the breakpoint could still trigger: look at the disassembly, confirm real work is done, or you don't know. And for whatever reason you failed to answer "was it debug or release". — Yakk - Adam Nevraumont, Aug 27 '18 at 12:56

Deduplicator · Accepted Answer · 2020-11-24T00:26:01.783

Well, the as-if-rule always allows evaluation at runtime. However insane (and insanely complex) doing so might be.

Best shot to force your compiler to do it at compile-time, pass it through a template-argument:

A bit of setup:

template <auto x>
using make_integral_constant = std::integral_constant<decltype(x), x>;

template <auto x>
inline constexpr auto want_static = make_integral_constant<x>::value;

And use it like:

foo( want_static<CompileTimeHash( "aha" )> );

It works even without optimization, because unless you use an interpreter, doing it at runtime instead is too complex to do for no good reason.

Assigning to a constexpr-variable should also work. But it is actually easier to not evaluate at compile-time, so without optimization that happens anyway.

foo( []{ constexpr auto r = CompileTimeHash( "aha" ); return r; }() );

Update: In C++ 20 a new keyword consteval can be used to define an immediate function, which will always be evaluated at compile-time. constinit can be used to force initialization with constexpr rules without making it constexpr.

consteval auto immediate(auto x) { return x; }

max66 · Answer 2 · 2018-08-25T11:23:49.397

-1

If I understand correctly, your CompileTimeHash() return a int.

So what about

foo( sizeof(char[CompileTimeHash( "aha" )]) );

?

If CompileTimeHash() return only positive numbers, obviously.

If CompileTimeHash() return non negative numbers (positive numbers or zero), you can solve the zero problem (not acceptable as size for a C-style array) adding (inside) and subtracting (outside) 1

I mean

foo( sizeof(char[CompileTimeHash( "aha" )+1])-1u );

edited Aug 25 '18 at 11:23

answered Aug 24 '18 at 13:33

max66

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@old123987 - ehmm... no; if `CompileTimeHash()` return negative numbers or zero this can't work; but with positive numbers... – max66 Aug 24 '18 at 13:38
1

Even if it were an unsigned int rather than an int - this is a convoluted solution. Baroque, confusing, likely to be undone or otherwise abused by future maintainers. -1. – einpoklum Aug 24 '18 at 15:38
Even if it were an unsigned int, my solution is problematic for the zero case; I disagree with you for the following (convoluted, baroque, confusing...) but thanks for the explanation of your down-vote. – max66 Aug 24 '18 at 16:43

constexpr does not work/apply inside function call

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