为什么允许某些非常量表达式作为 constexpr 逗号运算符的操作数？

Question

考虑一个简单的例子：

int foo() {
    return 3;
}

template <int>
struct Bar {};

int a;

int main() {
    int b;
    //Bar<((void)foo(), 1)> bar1;  //case 1. compilation error as expected
    Bar<((void)a, 2)> bar2;        //case 2. no error (long shot but `a' has a linkage so maybe expected)
    Bar<((void)b, 3)> bar3;        //case 3. no error ? (`b' does not have linkage) 
    (void)bar2;
    (void)bar3;
}

我会说这是一个错误，但最新的 [clang] and [gcc] 都接受了代码，所以也许我遗漏了一些使代码有效的相关标准规则？

Answer 1

左值到右值的转换不适用于逗号运算符的第一个参数，除非它是可变的。因此，(void)a, 2和(void)b, 3是常量表达式。

见[expr.comma]/1

... the left expression is a discarded-value expression ...

和[expr]/12

... The lvalue-to-rvalue conversion is applied [to a discarded-value expression] if and only if the expression is a glvalue of volatile-qualified type and it is one of the following: ...

为什么允许某些非常量表达式作为 constexpr 逗号运算符的操作数？

Why are some non-constant expressions allowed as operands of a constexpr comma operator?

c++

c++11

language-lawyer

constexpr

comma-operator