删除转换运算符

Question

有没有办法禁用转换运算符？将它们标记为“=删除”会搞乱其他事情。

考虑以下代码：

class Foo
{
public:

    Foo() :mValue(0) {}
    ~Foo() = default;
    Foo(int64_t v) { mValue = v; }
    Foo(const Foo& src) = default;

    bool operator==(const Foo& rhs) { return mValue == rhs.mValue; }

    /* after commenting these lines the code will compile */
    operator int() const = delete;
    operator int64_t() const = delete;

private:
    int64_t mValue;
};

int main()
{
    Foo foo1(5);
    Foo foo2(10);
    bool b1 = (foo1 == foo2);
    bool b2 = (foo1 == 5);
}

这不会编译，因为 gcc 抱怨 == 运算符不明确：

test.cc: In function ‘int main()’:
test.cc:25:21: error: ambiguous overload for ‘operator==’ (operand types are ‘Foo’ and ‘int’)
     bool b2 = (foo1 == 5);
                     ^
test.cc:25:21: note: candidates are:
test.cc:25:21: note: operator==(int, int) <built-in>
test.cc:25:21: note: operator==(int64_t {aka long int}, int) <built-in>
test.cc:10:10: note: bool Foo::operator==(const Foo&)
     bool operator==(const Foo& rhs) { return mValue == rhs.mValue; }
          ^

但是，在注释转换运算符之后，代码将编译并且运行很好。

第一个问题是：为什么删除的转换运算符会为==运算符造成歧义？我认为他们应该禁用隐式 Foo -> int 转换，但它们似乎会影响 int -> Foo 转换，这对我来说听起来不合逻辑。

第二个：有没有办法标记转换运算符被删除？是的，通过不声明它们 - 但我正在寻找一种方式，将来任何人都会看到这些转换被设计禁用。

Answer 1

Any use of a deleted function is ill-formed (the program will not compile).

If the function is overloaded, overload resolution takes place first, and the program is only ill-formed if the deleted function was selected.

在你的情况下，程序无法select转换，因为你有 3 个变体

int -> Foo
Foo -> int
Foo -> int64

第二个问题：你可以保持原样，但始终对 int

使用显式转换

bool b2 = (foo1 == Foo(5));

Answer 2

我认为问题的关键在于：

[dcl.fct.def.delete]:

A program that refers to a deleted function implicitly or explicitly, other than to declare it, is ill-formed.
...
A deleted function is implicitly an inline function ([dcl.inline]).

[class.member.lookup/4]:

If C contains a declaration of the name f, the declaration set contains every declaration of f declared in C that satisfies the requirements of the language construct in which the lookup occurs.

即使你 delete 函数，你仍然声明它。并且声明的函数将参与重载决议。只有当它是已解析的重载时，编译器才会检查它是否被删除。

在您的情况下，当这些函数声明存在时，存在明显的歧义。

删除转换运算符

Deleting conversion operators

c++

operators

implicit-conversion

deleted-functions