c++ - 三元运算符隐式转换为基类

Question

考虑这段代码:

struct Base
{
    int x;
};

struct Bar : Base
{
    int y;
};

struct Foo : Base
{
    int z;
};

Bar* bar = new Bar;
Foo* foo = new Foo;

Base* returnBase()
{
    Base* obj = !bar ? foo : bar;
    return obj;
}

int main() {
    returnBase();
    return 0;
}

这在 Clang 或 GCC 下不起作用，给我:

error: conditional expression between distinct pointer types ‘Foo*’ and ‘Bar*’ lacks a cast Base* obj = !bar ? foo : bar;

这意味着它要编译我必须将代码更改为:

Base* obj = !bar ? static_cast<Base*>(foo) : bar;

由于存在对 Base* 的隐式强制转换，是什么阻止了编译器这样做？

换句话说，为什么 Base* obj = foo; 可以在没有强制转换的情况下工作，但使用 ?: 运算符却不行？是不是因为不清楚我要使用 Base 部分吗？

Answer 1

引用 C++ 标准草案 N4296，第 5.16 节 条件运算符，第 6.3 段:

• One or both of the second and third operands have pointer type; pointer conversions (4.10) and qualification conversions (4.4) are performed to bring them to their composite pointer type (Clause 5). The result is of the composite pointer type.

第 5 节 表达式，第 13.8 和 13.9 段:

The composite pointer type of two operands p1 and p2 having types T1 and T2, respectively, where at least one is a pointer or pointer to member type or std::nullptr_t, is:

• if T1 and T2 are similar types (4.4), the cv-combined type of T1 and T2;
• otherwise, a program that necessitates the determination of a composite pointer type is ill-formed.

注意:我在此处复制 5/13.8 只是为了向您展示它没有命中。实际生效的是 5/13.9，“程序格式错误”。

以及第 4.10 节 指针转换，第 3 段:

A prvalue of type “pointer to cv D”, where D is a class type, can be converted to a prvalue of type “pointer to cv B”, where B is a base class (Clause 10) of D. If B is an inaccessible (Clause 11) or ambiguous (10.2) base class of D, a program that necessitates this conversion is ill-formed. The result of the conversion is a pointer to the base class subobject of the derived class object. The null pointer value is converted to the null pointer value of the destination type.

因此，Foo 和 Bar 都派生自同一个基类并不重要(完全)。重要的是指向 Foo 的指针和指向 Bar 的指针不能相互转换(没有继承关系)。