Question

我正在创建一个自定义对象的向量，并按如下方式调用assign：

class myClass
{
        public:
                myClass() { cout<<"MyClass def const"<<endl; }
                myClass(const myClass &mclass) {cout<<"Default const"<<endl;}
                myClass& operator=(myClass &mclass) { cout<<"called overloaded = operator"<<endl; return mclass; }
};


int main()
{
        myClass m;
        cout<<"orginal object:"<<endl;
        vector<myClass> vec1,vec2,vec3;
        vec1.assign(10,m);
        return 0;
}

相关编译错误：

/usr/lib/gcc/i386-redhat-linux/4.3.2/../../../../include/c++/4.3.2/bits/stl_vector.h:344:   instantiated from ‘void std::vector<_Tp, _Alloc>::assign(size_t, const _Tp&) [with _Tp = myClass, _Alloc = std::allocator<myClass>]’
test.cpp:52:   instantiated from here
/usr/lib/gcc/i386-redhat-linux/4.3.2/../../../../include/c++/4.3.2/bits/stl_algobase.h:686: error: no match for ‘operator=’ in ‘* __first = __value’
test.cpp:43: note: candidates are: myClass& myClass::operator=(myClass&)

我不知道我错过了什么，因为我重载了=运算符。

编辑：

我的签名似乎错误如下所述。据我所知，有效签名是：

(1) MyClass& operator=( const MyClass& rhs ); 
(2) MyClass& operator=( MyClass& rhs ); 
(3) MyClass& operator=( MyClass rhs ); 
(4) const MyClass& operator=( const MyClass& rhs ); 
(5) const MyClass& operator=( MyClass& rhs ); 
(6) const MyClass& operator=( MyClass rhs ); 
(7) MyClass operator=( const MyClass& rhs ); 
(8) MyClass operator=( MyClass& rhs ); 
(9) MyClass operator=( MyClass rhs );

将参数作为const引用定义的限制在哪里？

Answer 1

赋值运算符的参数必须为const，例如

myClass& operator=(const myClass &mclass)

自定义对象的向量：分配编译失败

1 个答案: