什么决定了 constexpr function 是否是常量表达式？

Question

(compiler used is gcc with c++17 as far as I know (difficult to find this in visual studio))

#include <iostream>

using namespace std;

void increment( int& v )
{
    ++v;
}

int constexpr f()
{
    int v = 0;
    increment( v );
    return v;
}

int main( )
{
    cout << f( ) << '\n';
}

The above code gives the error on compile:

constexpr function 'f' cannot result in a constant expression.

As I understand it this is because the function increment is not a constexpr. What confuses me is that the following code compiles fine:

#include <iostream>

using namespace std;

void increment( int& v )
{
    ++v;
}

int constexpr f()
{
    int v = 0;
    for( int i = 0; i < 1; ++i )
    {
        increment( v );
    }   
    return v;
}

int main( )
{
    cout << f( ) << '\n';
}

This code is functionally the same and it does compile, even though increment is still not a constexpr. I don't understand how it's possible that a for-loop through the range [0, 1) causes the compiler to realize that the function f actually is a constexpr.

If anyone can give some insights on constexpr in c++ and this apparent inconsistency, I'd greatly appreciate it.

Answer 1

Both programs are "ill-formed no diagnostic required", per [dcl.constexpr]/6 :

For a constexpr function or constexpr constructor that is neither defaulted nor a template, if no argument values exist such that an invocation of the function or constructor could be an evaluated subexpression of a core constant expression, or, for a constructor, an evaluated subexpression of the initialization full-expression of some constant-initialized object ( [basic.start.static] ), the program is ill-formed, no diagnostic required.

It's a bit strange that gcc just fails to notice the issue with the second program, but it's still conforming.

Note a diagnostic would be required if f were used in a context that actually requires a constant expression, for example constexpr int n = f(); .

Some things are never permitted in a constexpr function. These do require a diagnostic (typically an error message), even if the function is never used in a constant expression - see cigien's answer . But the programs in the question don't violate any of these stricter rules.

Answer 2

Since you're not calling f in a constant expression, your question is asking if the compiler is required to diagnose that f can't be called in a constant expression, based solely on its definition .

The requirements on the definition of a constexpr function are enumerated here :

The definition of a constexpr function shall satisfy the following requirements:

(3.1) its return type (if any) shall be a literal type;

(3.2) each of its parameter types shall be a literal type;

(3.3) it shall not be a coroutine;

(3.4) if the function is a constructor or destructor, its class shall not have any virtual base classes;

(3.5) its function-body shall not enclose

(3.5.1) a goto statement,

(3.5.2) an identifier label,

(3.5.3) a definition of a variable of non-literal type or of static or thread storage duration.

As can be seen, the definition of f does not violate any of the requirements in the list. So a compiler is conforming if it chooses not to diagnose this.

As pointed out in aschepler's answer , constexpr functions like f that can't be called in a constant expression, but are not diagnosable as such, are considered ill-formed-no-diagnostic-required.

Answer 3

You're not actually "calling" f at compile time.

if your main function included: static_assert(f() == 1, "f() returned 1"); I suspect you would get an "f() is not a constant expression" error.

Here's a related question

Answer 4

The standard requires that a constexpr function actually be evaluable at compile time for some set of parameters but not all. It does not require compilers to diagnose a constexpr function doing certain things which might be non-compile-time in some circumstances, or even whether such a function has such a set of parameters. This avoids them having to solve the halting problem.