一个神奇的bug-string构造函数触发

妖狐艹你老母 2023-01-21 10:19 5阅读 0赞

\#include < iostream >  
using namespace std;

template < class T >  
class node  
\{  
public:  
T info;  
node < T > \* link;  
public:  
node(const T& t = 0) :info(t), link(NULL)  
\{

\}  
\};

template < typename T >  
class myqueue  
\{  
int n\_size;  
public:  
myqueue() :head(NULL), tail(NULL), n\_size(0) \{\}  
~myqueue()  
\{  
clear();  
\}  
void push(const T& t) //向队伍增添成员，和链表一样  
\{  
node < T > \* temp = new node; //新  
if (!head)  
\{  
head = temp;  
tail = head;  
\}  
else  
\{  
temp->info = t; //装  
tail->link = temp; //连  
tail = temp;  
\}  
n\_size++;  
\}

void pop() //弹出队伍  
\{  
if (empty())  
\{  
return;  
\}  
node < T > \* temp = head;  
head = head->link;  
delete temp;

if (head == NULL)  
tail = NULL;

n\_size = 0;  
\}

int size()  
\{  
return n\_size;  
\}

void clear()  
\{  
node < T > \* temp;  
while (head)  
\{  
temp = head;  
head = head->link;  
delete temp;  
\}  
n\_size = 0;

bool empty() //判断队列是否是空队  
\{  
return n\_size == 0;  
\}

T& front() //查看队头，返回头部数据  
\{  
return head->info;  
\}

T& back() //查看队尾，返回尾部数据  
\{  
return tail->info;  
\}  
protected:  
node < T > \* head;  
node < T > \* tail;  
\};

int main()  
\{  
myqueue < int > q;  
int i = 0;  
int j = 0;  
int k = 0;

for (i = 0; i < 10; i++)  
q.push(i);

for (k = 0; !q.empty(); k++)  
\{  
if (k != q.front())  
\{  
cout << "Test case 1: Failed ..." << endl;  
return 1;  
\}  
cout << q.front();  
q.pop();  
\}  
cout << endl;  
cout << "Test case 1: passed" << endl;

// ---- 不太华丽的分割线-----------  
for (j = 0; j < 999; j++)  
\{  
q.push(j);  
\}  
q.clear();  
if (q.size() != 0)  
\{  
cout << "Test case 2: failed ..." << endl;  
return 1;  
\}  
cout << "Test case 2: passed..." << endl;

// ---- 不太华丽的分割线-----------  
myqueue < string > q\_s;  
int loop\_time = 9999;  
for (i = 1; i < loop\_time; i++)  
\{  
q\_s.push(string("string:") += i);  
\}

for (i = 0; i < loop\_time + 100; i++)  
\{  
q\_s.pop();  
\}  
if (q\_s.size() != 0)  
\{  
cout << "Test case 3: failed ..." << endl;  
return 1;  
\}  
cout << "Test case 3: passed..." << endl;

// ---- 不太华丽的分割线-----------  
myqueue < int > q\_i;  
q\_i.clear();  
for (i = 0; i < loop\_time; i++)  
\{  
q\_i.push(i);  
\}  
q\_i.clear();  
if (q\_i.size() != 0)  
\{  
cout << "Test case 4: failed ..." << endl;  
return 1;  
\}  
cout << "Test case 4: passed..." << endl;  
\}

================================================================

这个代码会出错！错在 mylist < T > :: push\_back()函数

继续查，错在node的构造

继续查，错在string构造

继续查，string用0构造

继续查，原因在这个构造函数

template < class T >  
class node  
\{  
public:  
T info;  
node\* link;  
public:  
node(const T& t = 0) :info(t), link(NULL)  
\{

\}  
\};

string用0构造，相当于传入const char\* s=NULL, 完蛋！

怎么办，有个偷懒方法，修改构造函数为

node(const T& t = T()) :info(t), link(NULL)  
\{

这样，当无参数传入构造node, 相当于构造一个临时无名变量string(), 这样就不出错啦！！！