Question

在GCC后期版本中。当一个人有多个带引用或指针的构造函数时，如何（如果可以的话）消除'0'或'NULL'的歧义？

即：

class XXX
{
public:
    XXX(const XXX &tocopy);
    XXX(const char *fromAString);
    XXX(const AnotherThing *otherThing);

    operator=(const XXX &tocopy);
    operator=(const char *fromAString);
    operator=(const AnotherThing *otherThing);
};

// nice not to have to cast when setting to NULL for 
// things like smart pointers and strings. Or items that can be initialized from 
// several types of objects and setting to null means "clear"

XXX anXXX = NULL;
anXXX = 0;

// In MSC one can have an 
//    XXX(const int &nullItem) { DEBUG_ASSERT(!nullItem); setnull(); } 
// and the overloads would be disambiguated.  GCC will cause a const int to conflict
// with pointer types.

Answer 1

C ++有类型系统，因此变量具有类型，编译器使用这些类型来执行重载解析：

const char * p = 0;
const AnotherThing * q = 0;

XXX a(p), b(q); // uses the respective constructors for the static type of p, q

如果由于您没有使用所需的指针类型之一而导致重载不明确，则会出现错误：

XXX c(0); // error: ambiguous

可以为多个构造函数消除歧义0（NULL）吗？和赋值运算符？

1 个答案: