三元运算符隐式转换为基类

问题描述：

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：

This doesn't work under Clang or GCC, giving me :

错误：不同指针类型'Foo *'
和'Bar *'之间的条件表达式缺少强制转换的Base * obj =！bar？ foo：bar;

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

这意味着要编译，我必须将代码更改为：

Which means for it to compile I have to change the code to :

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

因为隐式强制转换为 Base * 存在，是什么阻止了编译器这样做？

Since an implicit cast to a Base* exists, what is preventing the compiler from doing so?

换句话说，为什么 Base * obj = foo; 无需强制转换但可以使用？：运算符来工作吗？是因为不清楚我是否要使用 Base 部分？

In other words, why does Base* obj = foo; work without a cast but using the ?: operator doesn't? Is it because it's not clear that I want to use the Base part?

答

引用C ++标准草案N4296第5.16节有条件的运算符第6.3段：

Quoting from C++ standard draft N4296, Section 5.16 Conditional operator, Paragraph 6.3:

第二和第三操作数中的一个或两个具有指针类型；进行了指针转换（4.10）和
资格转换（4.4），以将其转换为复合指针类型（第5条）。
结果是复合指针类型。

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：

Section 5 Expressions, Paragraph 13.8 and 13.9:

两个分别具有T1和T2类型的操作数p1和p2的复合指针类型，在
处，至少有一个是指向成员类型或std :: nullptr_t的指针或指针，是：

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:

如果T1和T2是类似的类型（4.4），T1和T2的cv组合类型；

否则，需要确定复合指针类型的程序格式错误。

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：我在这里复制5 / 13.8只是为了向您展示它没有打。实际上是5 / 13.9，程序格式错误。

Note: I copied 5/13.8 here just to show you that it doesn't hit. What's actually in effect is 5/13.9, "the program is ill-formed".

第4.10节指针转换，第3段：

And Section 4.10 Pointer conversions, Paragraph 3:

类型为 pointer to cv D的prvalue（其中D是类类型）可以转换为类型为 pointer $的prvalue b $ b到Cv B，其中B是D的基类（第10条）。如果B是D的不可访问（第11条）或模棱两可（10.2）
，则该程序需要进行此转换格式不正确。转换的结果是指向派生类对象的基类子对象的
指针。空指针值将转换为目标类型的
空指针值。

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，没关系（完全）Foo和Bar都源自一个相同的基类。唯一重要的是，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）。

So, it doesn't matter (at all) that both Foo and Bar are derived from one same base class. It only matters that a pointer to Foo and a pointer to Bar are not convertible to each other (no inheritance relationship).

三元运算符隐式转换为基类

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