向量自定義對象：指定編譯失敗

Question

我創建的自定義對象的矢量和呼叫分配如下：

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; 
} 

相關的編譯錯誤：

/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&) 

我不知道我是什麼因爲我重載了=運算符而錯過了。

編輯：

看來我的簽名是錯誤的，如下更正。據我瞭解，有效簽名是：

(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); 

限制傳遞參數作爲const引用定義的限制在哪裏？

Answer 1

賦值運算符的參數必須是const，例如，

myClass& operator=(const myClass &mclass)