为什么 consteval 函数允许未定义的行为？

Question

C++ 中有一个非常整洁的常量表达式属性：它们的计算不能有未定义的行为(7.7.4.7)：

An expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine ([intro.execution]), would evaluate one of the following:

...

an operation that would have undefined behavior as specified in [intro] through [cpp] of this document [ Note: including, for example, signed integer overflow ([expr.prop]), certain pointer arithmetic ([expr.add]), division by zero, or certain shift operations — end note ] ;

试图将 13! 的值存储在 constexpr int 中确实 yields a nice compile error:

constexpr int f(int n) 
{
    int r = n--;
    for (; n > 1; --n) r *= n;
    return r;
}

int main() 
{
    constexpr int x = f(13);
    return x;
}

输出：

9:19: error: constexpr variable 'x' must be initialized by a constant expression
    constexpr int x = f(13);
                  ^   ~~~~~
4:26: note: value 3113510400 is outside the range of representable values of type 'int'
    for (; n > 1; --n) r *= n;
                         ^
9:23: note: in call to 'f(3)'
    constexpr int x = f(13);
                      ^
1 error generated.

（顺便说一句，为什么错误说“调用 'f(3)'”，而它是对 f(13) 的调用？..）

然后，我从 x 中删除 constexpr，但将 f 设为 consteval。根据 the docs:

consteval - specifies that a function is an immediate function, that is, every call to the function must produce a compile-time constant

我预计这样的程序会再次导致编译错误。但是，the program compiles and runs with UB.

这是为什么？

UPD: 评论者认为这是一个编译器错误。我举报了：https://bugs.llvm.org/show_bug.cgi?id=43714

Answer 1

这是一个编译器错误。或者，更准确地说，这是一个“未充分实现”的功能（参见 the comment in bugzilla）：

Yup - seems consteval isn't implemented yet, according to: https://clang.llvm.org/cxx_status.html

(the keyword's probably been added but not the actual implementation support)

为什么 consteval 函数允许未定义的行为？

Why does a consteval function allow undefined behavior?

c++

undefined-behavior

c++20

consteval