C++ 中的常量表达式有一个非常简洁的属性:它们的求值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 ] ;
确实尝试将 13! 的值存储在 constexpr int 中 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
我确实希望这样的程序会再次导致编译错误。但相反，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
最佳答案
这是一个编译器错误。或者，更准确地说，这是一个“未实现”的功能(请参阅 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)
关于c++ - 为什么保守函数允许未定义的行为？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/58454657/