重载的赋值运算符没有被调用

发布于 2024-09-12 14:41:25 字数 486 浏览 2 评论 0原文

我编写了类 perform 的重载赋值运算符，复制所有变量值。例如：在 Exp.cpp 中，

class perform
{
    LOG *ptr;
int a;
//constructor
//destructor
perform operator=(const perform & rhs){

   ptr = rhs.ptr; a=rhs.s;
return * this;}
};

在另一个类 output 中，我声明了一个 abc 的指针。

perform * ptr = StatCol::CreateCol(frm);
abc = ptr; //this line should invoke assignment overloaded.
           //but in my case it's not invoked.

原文

I have written a overloaded assignment operator of class perform copying all the variable values.
For ex :in Exp.cpp

class perform
{
    LOG *ptr;
int a;
//constructor
//destructor
perform operator=(const perform & rhs){

   ptr = rhs.ptr; a=rhs.s;
return * this;}
};

In another class output, I have declared a pointer for abc.

perform * ptr = StatCol::CreateCol(frm);
abc = ptr; //this line should invoke assignment overloaded.
           //but in my case it's not invoked.

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椵侞 2024-09-19 14:41:25

假设 abc 是一个 Perform 对象，您需要取消引用您正在分配的指针：

abc = * ptr;

如果 abc 本身是一个指针，那么您将无法执行您所要求的操作 - 您无法在 LHS 是指针的情况下重载分配。您必须取消引用两个指针：

* abc = * ptr;

Assuming abc is a Perform object, you need to dereference the pointer you are assigning:

abc = * ptr;

If abc itself is a pointer, then you can't do what you are asking - you can't overload assignment where the LHS is a pointer. You would have to dereference both pointers:

* abc = * ptr;

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十二 2024-09-19 14:41:25

此外，通过引用返回更安全，从而避免调用复制构造函数。

    const perform& operator=(const perform & rhs){

     if (this != &rhs)
     {
       ptr = rhs.ptr; a=rhs.s;
     }
     return * this;
   }

Also, it is safer to return via reference thus avoiding a copy constructor being invoked.

    const perform& operator=(const perform & rhs){

     if (this != &rhs)
     {
       ptr = rhs.ptr; a=rhs.s;
     }
     return * this;
   }

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望喜 2024-09-19 14:41:25

Custom assignment operator works only with user defined types so do like this:

perform p1,p2; 
p1 = p2;
perform *p = &p2;
p1 = *p;  


You can't override assignment of built in types(int , char etc.).

perform *p1,*p2; 
p1 = p2;

It simply copies the address of p2 to p1.

Custom assignment operator works only with user defined types so do like this:

perform p1,p2; 
p1 = p2;
perform *p = &p2;
p1 = *p;  


You can't override assignment of built in types(int , char etc.).

perform *p1,*p2; 
p1 = p2;

It simply copies the address of p2 to p1.

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