在 C++20 常量表达式中是否允许比较动态分配对象的地址？

Question

C++20 允许在 constexpr 函数中进行堆分配，只要内存不泄漏即可。然而，GCC 和 Clang 对于比较两个动态分配对象的地址是否是常量表达式存在分歧。

以下代码片段可以用 Clang 编译，但不能用 gcc 编译。

constexpr bool foo() {
    int* a = new int(4);
    int* b = new int(4);
    bool result = a == b;
    delete a;
    delete b;
    return result;
}

constexpr bool x = foo(); // GCC:  error: '(((int*)(& heap deleted)) == ((int*)(& heap deleted)))' is not a constant expression

以下在两个编译器上都可以正常工作

constexpr bool foo2() {
    int a = 4;
    int b = 5;
    bool result = &a ==  &b;

    return result;
}

constexpr bool x = foo2();

我假设为了正确删除动态对象，编译器必须知道指针是否指向相同的对象，所以我假设这是一个 GCC 错误（或尚未完全实现） .谁能证实这个假设？还是我错了？

实例here.

编辑：奇怪的是，当我通过提供的 link 打开实时示例时，它突然也在 gcc 上编译。但是如果我将它复制粘贴到一个新的编译器资源管理器实例中，它会再次失败。或者，如果我多次重新加载它，它会每隔两次失败一次并每隔两次编译一次...

Answer 1

这是一个 gcc 错误 (#85428)。

[expr.const]/5 that would cause evaluation of a == b to fail to be a constant expression. The only one there in which there is any question would be the one about undefined behavior. So we could go look at [expr.eq] 中没有关于指针比较的内容：

If at least one of the operands is a pointer, pointer conversions, function pointer conversions, and qualification conversions are performed on both operands to bring them to their composite pointer type. Comparing pointers is defined as follows:

If one pointer represents the address of a complete object, and another pointer represents the address one past the last element of a different complete object, the result of the comparison is unspecified.

Otherwise, if the pointers are both null, both point to the same function, or both represent the same address, they compare equal.

Otherwise, the pointers compare unequal.

两个指针都代表了不同的完整对象的地址，都不为null，所以就落入第三点，只是指针比较不相等。这里没有未定义或未指定的行为。

a == b 应该只产生 false.

在 C++20 常量表达式中是否允许比较动态分配对象的地址？

Is comparing addresses of dynamically allocated objects allowed in C++20 constant expression?

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