我正在学习使用链表实现Stack。这是节点类:
class StudentInfo {
public:
string id, name, course;
double GPA;
StudentInfo *next;
};
这是Stack类:
class StackLinkedList {
public:
StudentInfo* top; //pointer to point to the top node
int size; //variable to keep the size of the stack
//constructor
StackLinkedList() {
this->size = 0;
this->top = NULL;
}
//destructor
~StackLinkedList() {
StudentInfo *current = top;
while (top) {
current = current->next;
delete top;
top = current;
}
}
//to add item into stack - push on top
void push(StudentInfo *newStudent) {
if (!top) {
top = newStudent;
return;
}
newStudent->next = top;
top = newStudent;
size++;
}
void main() {
StudentInfo s1("phi", "123", "computer science", 4.0);
StudentInfo s2("abc", "123", "software engineer", 4.0);
StudentInfo s3("zxc", "123", "business management", 4.0);
StackLinkedList list;
StudentInfo *ptr;
ptr = &s1;
list.push(ptr);
ptr = &s2;
list.push(ptr);
ptr = &s3;
list.push(ptr);
};
当我尝试在push()和printAll()上运行单元测试时,一切正常。但是,在调用了析构函数()之后,出现错误Debug Assertion Failed ... is_block_type_valid(header-> _block_use)。并且调试器在delete top;
//destructor
~StackLinkedList() {
StudentInfo *current = top;
while (top) {
current = current->next;
delete top; //here
top = current;
}
}
如果我将top = NULL;
放在delete top;
之前,则错误消失了。所以,我对top = NULL;
语句有点混淆。
编辑:NodeType的构造函数
StudentInfo(string id, string name, string course, double gpa) {
this->id = id; this->name = name; this->course = course; this->GPA = gpa; this->next = NULL;
}
答案 0 :(得分:4)
您通过尝试自动存储时间的delete
个对象来调用未定义的行为。
int main() {
StudentInfo s1("phi", "123", "computer science", 4.0);
StudentInfo s2("abc", "123", "software engineer", 4.0);
StudentInfo s3("zxc", "123", "business management", 4.0);
StackLinkedList list;
StudentInfo *ptr;
ptr = &s1;
list.push(ptr);
ptr = &s2;
list.push(ptr);
ptr = &s3;
list.push(ptr);
};
如您所见,s1
,s2
,s3
是自动存储持续时间的对象(也就是说,编译器会在最后自动调用它们的析构函数他们的一生)。
然而,你将它们的地址传递给list
,其破坏者deletes
在其链接列表详细信息中的所有指针都会被破坏....永远不要在指针上调用delete
使用new
创建的对象。
一些补充说明:
void main()
在C ++中是非法的。你使用的是较旧的编译器吗? .. std::forward_list
使用分配器在内部管理其节点的分配。我建议你重新设计StackLinkedList
以在内部管理它的节点,这样客户就永远不会为生命而烦恼。答案 1 :(得分:1)
对于初学者,您不会初始化next
类型对象的数据成员StudentInfo
。
因此依赖于列表中最后一个节点等于nullptr
的所有代码都将调用未定义的行为。
此外,您不能对不是使用operator new创建的对象使用operator delete。
所以代替陈述
StudentInfo s1("phi", "123", "computer science", 4.0);
StudentInfo s2("abc", "123", "software engineer", 4.0);
StudentInfo s3("zxc", "123", "business management", 4.0);
你至少应该写(我假设StudentInfo
是一个聚合。如果该类有一个构造函数,那么声明它就像
StudentInfo( const string &id,
const string &name,
const string &course,
double gpa,
StudentInfo *next = nullptr )
{
this->id = id; this->name = name; this->course = course; this->GPA = gpa; this->next = next;
}
) StudentInfo * s1 =新的StudentInfo {" phi"," 123","计算机科学",4.0,nullptr}; StudentInfo * s2 =新的StudentInfo {" abc"," 123","软件工程师",4.0,nullptr}; StudentInfo * s3 =新的StudentInfo {" zxc"," 123","业务管理",4.0,nullptr};
然后
list.push(s1);
list.push(s2);
list.push(s3);