所以,我有一个令牌类: Token.h
class Token {
std::string name; // token name
int frequency;//frequency
Vector lines;//lines where the token is present
public:
//explanations for the methods in the Token.cpp
Token(std::string tokenname, int linenumber);
virtual ~Token();
const Vector getLines() const;
};
#endif /* TOKEN_H_ */
令牌cpp
Token::Token(string tokenname, int linenumber) {
// TODO Auto-generated constructor stub
name = tokenname;
frequency=1;
lines.push_back(linenumber);
}
Token::~Token() {
// TODO Auto-generated destructor stub
}
std::string Token::getName() const{
return name;
}
int Token::getFrequency() const{
return frequency;
}
const Vector Token::getLines() const{
const Vector vec = lines;
return lines;
}
当我将它传递给list class
的insert方法时,程序失败了class List {
private:
class Node {
public:
Token data;
Node* next;
Node(const Token &dataItem, Node* nextptr);
~Node();
};
Node* first;
int length;
public:
List();
virtual ~List();
void insert(const Token &t);
};
List.cpp:
List::Node::Node(const Token &dataItem, Node* nextptr): data(dataItem), next(nextptr){
}
List::Node::~Node(){
cout<<"dead"<<endl;
}
List::List() {
// TODO Auto-generated constructor stub
length = 0;
first = nullptr;
}
List::~List() {
// TODO Auto-generated destructor stub
Node* temp = first;
Node* newtmp;
while(temp->next != nullptr){
newtmp = temp->next;
delete temp;
temp = newtmp;
}
}
const int List::size(){
return length;
}
void List::insert (const Token &t){
Vector dammit = t.getLines();
}
我发现插入中的哪一行(Vector dammit = t.getLines()),所以我就这样离开了。 它给了我这个错误信息:
双重免费或损坏(fasttop):0x0000000000c34040 ***
如果您想要运行,请点击主文件中的内容:
int main() {
// cout<<"tokens are here"<<endl;
//
Token hit("aca", 1);
Token hit2("ui", 2);
Token hit1("111", 3);
List list;
list.insert(hit);
list.insert(hit2);
list.insert(hit1);
}
Vector class:
class Vector {
int* store;
int capacity;
int next_index;
public:
Vector();
Vector(int initial_size);
Vector(const Vector &v);
virtual ~Vector();
void push_back(int item);
int pop_back();
const int size() const;
void resize();
void operator =(const Vector &v);
int& operator[] (int k);
const int& operator[] (int k) const;
friend std::ostream& operator<<(std::ostream& os, const Vector& v);
};
Vector::Vector() {
// TODO Auto-generated constructor stub
store = new int [1];
capacity = 1;
next_index = 0;
}
Vector::Vector(int initial_size){
store = new int [initial_size];
capacity = initial_size;
next_index = 0;
}
Vector::Vector(const Vector &v){
store = v.store;
capacity = v.capacity;
next_index = v.next_index;
}
Vector::~Vector() {
// TODO Auto-generated destructor stub
delete[] store;
}
void Vector::resize(){
std::cout<<"in resize"<<std::endl;
std::cout<<capacity<<std::endl;
int length = capacity;
capacity+=100;
int* tempArray;
tempArray = new int[capacity];
for (int i=0; i<length; i++){
tempArray[i] = store[i];
}
if (length>1)
delete[] store;
std::cout<<"finish re4size"<<std::endl;
store = tempArray;
}
void Vector::push_back(int item){
if(next_index >= capacity)
this->resize();
store[next_index] =item;
next_index++;
}
int Vector::pop_back(){
next_index = next_index-1;
int last = store[next_index];
return last;
}
void Vector::operator =(const Vector &v){
//delete[] store;
store = v.store;
capacity = v.capacity;
next_index = v.next_index;
}
const int Vector::size() const{
return next_index-1;
}
int& Vector::operator[] (int k){
//assert((k<next_index)&(k>=0));
return store[k];
}
const int& Vector::operator[] (int k) const{
//assert((k<next_index)&(k>=0));
return store[k];
}
ostream& operator<<(ostream& os, const Vector& v)
{
for(int i=0; i<=v.size(); i++){
os << v[i]<< ' ';
}
return os;
}
答案 0 :(得分:4)
在
Vector::Vector(const Vector &v){
store = v.store;
capacity = v.capacity;
next_index = v.next_index;
}
现在有两个向量指向同一个int* store;
在
void Vector::operator =(const Vector &v){
//delete[] store;
store = v.store;
capacity = v.capacity;
next_index = v.next_index;
}
你做同样的事情。
致电
const Vector Token::getLines() const{
const Vector vec = lines;
return lines;
}
vec = lines
使用复制构造函数。您现在拥有指向同一商店的vec和行。
您返回行的副本,这将再次触发复制构造函数。第三个对象现在指向商店。
当堆栈展开时,本地定义的vec
被销毁。 ~Vector
delete
商店。您现在有两个对象指向同一个已取消分配的商店。
KABOOM!一旦你尝试用其中任何一个向量做任何其他事情。看起来像返回的Vector的销毁首先命中并导致析构函数重新删除存储。
您需要为新商店分配存储空间,然后将源存储库的内容复制到=运算符和复制构造函数中的新存储区中。
Vector::Vector(const Vector &v){
capacity = v.capacity;
store=new int[capacity];
for (size_t index; index < capacity; index++)
{
store[index] = v.store[index];
}
next_index = v.next_index;
}
和
Vector & Vector::operator =(const Vector &v){
delete[] store;
capacity = v.capacity;
store=new int[capacity];
for (size_t index; index < capacity; index++)
{
store[index] = v.store[index];
}
next_index = v.next_index;
}
可以使用 std::copy
代替C ++ 11中的for循环。也可以使用古老的memcpy
,但这只是因为存储是原始数据类型。
在我编辑时,感谢Jarod42,还有一点点调整:
const Vector & Token::getLines() const{ //note the return of a reference. This avoids
// making a copy of lines unless the caller really
// wants a copy.
// const Vector vec = lines; don't need to do this. lines is const-ified by the
// const on the return type of the function
return lines;
}
答案 1 :(得分:1)
此错误与调用具有常量引用的方法无关,而是与函数getLines()
无关。例如,如果您使用hit1
并直接调用函数getLines()
,它将会崩溃。问题在于Token
如何具有堆栈分配的属性Vector
,该属性又具有int *
属性。这不一定是个问题,但根据您实现这些类的方式,它可能会导致内存冲突。
如果您想继续使用getLine()
并且无法使用<vector>
库,则可以将Token
的{{1}}属性更改为{{1}并相应地更改所有其他语法。还记得初始化指针lines
内存,否则它会崩溃。
但是,如果不必要,我宁愿使用尽可能少的动态分配内存。和其他用户说的那样,在Vector *
之前你应该有一个条件lines