重载比较运算符的使用>在 C++与 getter 函数结合使用
我正在努力解决有关二进制比较运算符 > 重载的问题。按照设计，它应该比较两张卡并返回 1（如果左侧参数较大）或 0（相反情况）。
下面是对该问题的简要描述：
class Card 包括变量 intsuit 和 int value 作为私有数据成员。我已将重载运算符函数声明如下：
int operator>(const Card& lhs, const Card& rhs);
因为它需要访问 Card 类的私有数据成员，它是在类声明中使用 friend 限定符声明的。
该功能本身已被确认可以按描述工作。真正的问题在于通过调用以下形式的“getter”函数来提供两个参数：
Card &Node::getCardRef() const{
Card& ref = *c;
return ref;
}
其中变量 c 的类型为 Card * 并指向一个有效对象类型为卡。另外，Node 类 的实例表示单链表中的一个节点。
按以下方式组合这两个函数会导致段错误（具体来说，用 gdb 术语“In Card &Node::getCardRef(): this = 0x0”）：
if (node.getCardRef() > node.getNext()->getCardRef()){
/* do wondrous stuff */
}
此外，在隔离时，Card &Node::getCardRef () 似乎产生了预期的结果。
I'm struggling with a problem concerning the overloading of the binary comparison operator >. By design, it is supposed to compare two cards and return either 1 (if the left-hand-side argument is bigger) or 0 (in the opposite case).
Here's a brief description of the problem:
class Card includes, among other stuff, the variables int suit and int value as private data members. I've declared the overloaded operator function as follows:
int operator>(const Card& lhs, const Card& rhs);
Because it needs to access private data members of class Card, it is declared with the friend qualifier in the class declaration.
The function itself is confirmed to work as described. The real problem lies with providing the two arguments by calling a 'getter' function of the following form:
Card &Node::getCardRef() const{
Card& ref = *c;
return ref;
}
where the variable c is of type Card * and points to a valid object of type Card. Also, an instance of class Node represents a node in a singly-linked list.
Combining the two functions in the following manner causes a segfault (specifically, in gdb terms "In Card &Node::getCardRef(): this = 0x0"):
if (node.getCardRef() > node.getNext()->getCardRef()){
/* do wondrous stuff */
}
Also, when isolated, Card &Node::getCardRef() seems to produce desired results.
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Node::getCardRef在该代码片段中被调用两次。第一次是.运算符的结果，因此我们可以合理地确定*其this将是有效的。对
Node::getCardRef的另一个调用是->运算符的结果。->的左侧完全有可能是 0（因此，this也将是 0）。node.getNext()很可能返回 0。单链表通常通过返回空指针来指示列表结束条件。我猜您正在将链接列表中的最后一项与不跟随它的空项进行比较。
*: we can be reasonably certain, but not 100% certain. It is possible that node contains a corrupt reference, or that a previous wild pointer has corrupted our local variables. In my experience, null pointers are much more likely than null references.
Node::getCardRefis called twice in that code fragment. The first time as a result of the.operator, so we can be reasonably certain* that itsthiswill be valid.The other call to
Node::getCardRefis the result of the->operator. It is entirely likely that the left-hand-side of the->is 0 (thus,thiswill be 0, also).It is very likely the case that
node.getNext()is returning 0. Singly-linked lists usually indicate the end-of-list condition by returning a null pointer.I guess that you are comparing the final item on your linked list with the null item that doesn't follow it.
*: we can be reasonably certain, but not 100% certain. It is possible that node contains a corrupt reference, or that a previous wild pointer has corrupted our local variables. In my experience, null pointers are much more likely than null references.