C++ constexpr function in return statement

Question

Why is a constexpr function no evaluated at compile time but in runtime in the return statement of main function?

It tried

template<int x>
constexpr int fac() {
    return fac<x - 1>() * x; 
} 

template<>
constexpr int fac<1>() {
    return 1; 
} 

int main() {
    const int x = fac<3>();
    return x;
}

and the result is

main:
        push    rbp
        mov     rbp, rsp
        mov     DWORD PTR [rbp-4], 6
        mov     eax, 6
        pop     rbp
        ret

with gcc 8.2. But when I call the function in the return statement

template<int x>
constexpr int fac() {
    return fac<x - 1>() * x; 
} 

template<>
constexpr int fac<1>() {
    return 1; 
} 

int main() {
    return fac<3>();
}

I get

int fac<1>():
        push    rbp
        mov     rbp, rsp
        mov     eax, 1
        pop     rbp
        ret
main:
        push    rbp
        mov     rbp, rsp
        call    int fac<3>()
        nop
        pop     rbp
        ret
int fac<2>():
        push    rbp
        mov     rbp, rsp
        call    int fac<1>()
        add     eax, eax
        pop     rbp
        ret
int fac<3>():
        push    rbp
        mov     rbp, rsp
        call    int fac<2>()
        mov     edx, eax
        mov     eax, edx
        add     eax, eax
        add     eax, edx
        pop     rbp
        ret

Why is the first code evaluated at compile time and the second at runtime?

Also I tried both snippets with clang 7.0.0 and they are evaluated at runtime. Why is this not valid constexpr for clang?

All evaluation was done in godbolt compiler explorer.

You should use -O3 for optimal behaviour, the question is meaningless without it. As posted the question is really about "why does the compiler without optimization do this thing" to which the answer is usually "to make debugging easier" — M.M
And both compilers in question do exactly what you would expect with optimizations enabled: godbolt.org/z/cCbFDX — Max Langhof
constexpr exists to let you write natural code for producing constant expressions in contexts that need them. It's not a "optimize this to heck always" hint to the compiler. It's best not to confuse it for one. — StoryTeller - Unslander Monica
@StoryTeller that comment helped me more than thousands books. I think you should make it an answer. I always thought constexpr is to tell the compiler "this should be evaluated at compile time", while I now understand that it is rather "this must be evaluatable at compile time". Suddenly the confusion about whether it does get evaluated at compile time or not feels much less severe. — 463035818_is_not_a_number
It's worth mentioning that in this particular example there is no need to pass x as a template parameter. An "ordinary" parameter will suffice here. So, you can use constexpr functions as a more readable replacement for some old style meta-programming-template-code that previously had to be used to generate compile time integers. — sebrockm

StoryTeller - Unslander Monica StoryTeller - Unslander Monica · Accepted Answer · 2019-01-14T13:08:49

A common misconception with regard to constexpr is that it means "this will be evaluated at compile time"¹.

It is not. constexpr was introduced to let us write natural code that may produce constant expressions in contexts that need them. It means "this must be evaluatable at compile time", which is what the compiler will check.

So if you wrote a constexpr function returning an int, you can use it to calculate a template argument, an initializer for a constexpr variable (also const if it's an integral type) or an array size. You can use the function to obtain natural, declarative, readable code instead of the old meta-programming tricks one needed to resort to in the past.

But a constexpr function is still a regular function. The constexpr specifier doesn't mean a compiler has² to optimize it to heck and do constant folding at compile time. It's best not to confuse it for such a hint.

_{¹ - Thanks user463035818 for the phrasing.
² - c++20 and consteval is a different story however :)}

C++ constexpr function in return statement

2 Answers