c++ - constexpr 函数中的复合赋值 : gcc vs. clang

Question

template<class A, class B> constexpr int f(A a, B b) {
    a /= b;
    return a;
}

constexpr int x = f(2, 2);   // a, b: int
constexpr int y = f(2., 2.); // a, b: double
constexpr int z = f(2, 2.);  // a: int, b: double //<-- BOOM!
constexpr int w = f(2., 2);  // a: double, b: int

int main() {}

代码未在 clang 中编译，它会产生以下诊断信息:

error: constexpr variable 'z' must be initialized by a constant expression

MSVC 崩溃(根据 godbolt )并且 gcc 工作正常。如果 a/= b 被简单地替换为 a = a/b 那么每个人都接受它。为什么？

谁是对的？似乎它与隐式缩小转换有关，但是为什么 a = a/b 有效？

Answer 1

如果我们查看复合赋值 [expr.ass]p7，这只是一个 clang 错误。它相当于 E1 只计算一次的赋值:

The behavior of an expression of the form E1 op= E2 is equivalent to E1 = E1 op E2 except that E1 is evaluated only once. In += and -=, E1 shall either have arithmetic type or be a pointer to a possibly cv-qualified completely-defined object type. In all other cases, E1 shall have arithmetic type.

如果我们查看 [dcl.constexpr]p3 中对常量表达式函数要求的限制我们对分配没有任何限制:

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

• (3.1) its return type shall be a literal type;
• (3.2) each of its parameter types shall be a literal type;
• (3.3) its function-body shall not contain.
• (3.3.1) an asm-definition,
• (3.3.2) a goto statement,
• (3.3.3) an identifier label ([stmt.label]),
• (3.3.4) a definition of a variable of non-literal type or of static or thread storage duration or for which no initialization is performed.
  [ Note: A function-body that is = delete or = default contains none of the above. — end note  ]

[expr.const] 中什么也没有为这种特定情况添加限制。

我离线联系了 Richard Smith，他同意这是一个错误并说:

Yes, it's a bug; that code is not correctly taking into account that the LHS could need conversion to floating point before the computation.

归档 clang bug report和 MSVC bug report