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C++的模板类中使用链表,却在一些环节报错说一些指针为空,但是这些地方都是限制过只有非空指针才能进的循环,没有理由会报错,求解。
#include<iostream>
using namespace std;
//链表节点
template<typename T>
struct List {
T num;
List* next;
};
//数组类
template<typename T>
class SString {
public:
template<typename T>
friend ostream& operator<<(ostream& out, const SString<T>& string);
SString() {
head = NULL;
}
SString(const SString& string) {
List<T> *t;
for (t = string.head; t != NULL; t = t->next) {
*this + t->num;
}
}
~SString() {
List<T> *t1, *t2;
t1 = head;
while (t1 != NULL) {
t2 = t1->next;
delete t1;
t1 = t2;
}
}
SString<T>& operator=(const SString<T>& string);
SString<T>& operator+(T x);
SString<T>& operator-(T x);
SString<T> operator+(const SString<T>& string)const;
SString<T> operator-(const SString<T>& string)const;
SString<T> operator*(const SString<T>& string)const;
private:
List<T> *head;
};
template<typename T>
SString<T>& SString<T>::operator+(T x) {
List<T> *t1, *t2, *t3;
t2 = new List<T>();
t2->num = x;
t2->next = NULL;
if (head == NULL) {
head = t2;
}else{
for (t1 = head; t1 != NULL; t1 = t1->next) {
t3 = t1;
}
t3->next = t2;
}
return *this;
}
template<typename T>
SString<T>& SString<T>::operator-(T x) {
List<T> *t1, *t2;
for (t1 = head; t1 != NULL; t1 = t1->next) {
if (t1->num == x) {
t2->next = t1->next;
delete t1;
t1 = t2->next;
}
t2 = t1;
}
return *this;
}
template<typename T>
SString<T> SString<T>::operator+(const SString<T>& string) const
{
List<T> *t;
SString<T> s(*this);
for (t = string.head; t != NULL; t = t->next) {
cout << t->num;
s.operator+(t->num);
}
return s;
}
template<typename T>
SString<T> SString<T>::operator-(const SString<T> & string) const
{
SString<T> s(*this);
List<T> *t;
for (t = string.head; t != NULL; t = t->next) {
s - t->num;
}
return s;
}
template<typename T>
SString<T> SString<T>::operator*(const SString<T> & string) const
{
SString<T> s;
s = *this - string;
s = *this - s;
return s;
}
template<typename T>
ostream& operator<<(ostream& out, const SString<T>& string)
{
out << "{";
List<T> *t;
for (t = string.head; t != NULL; t = t->next) {
out << t->num;
if (t->next != NULL) {
out << ",";
}
}
out << "}";
return out;
}
template<typename T>
SString<T>& SString<T>::operator=(const SString<T>& string) {
if (this == &string) {
return *this;
}
List<T> *t;
for (t = string.head; t != NULL; t = t->next) {
*this + t->num;
cout << t->num;
}
return *this;
}
int main() {
SString<int> s1, s2, s3, s4, s5;
for (int i = 0; i < 3; i++) {
int num;
cin >> num;
s1 = s1 + num;
}
cout << s1 << endl;
for (int i = 0; i < 3; i++) {
int num;
cin >> num;
s2 = s2 + num;
}
cout << s2 << endl;
cout << "ss";
(s1 + s2);
cout << "ss";
//cout << s3 << endl;
//s4 = s1 - s2;
//cout << s4 << endl;
//s5 = s1*s2;
//cout << s5 << endl;
return 0;
}
黄舟2017-04-17 13:47:08
代码的其中一段确实在gcc中编译不过,类型名字换一下就行了,然后就是运行没有问题,希望你把问题补充的再具体些。。比如出现错误的截图
template<typename X>
friend ostream& operator<<(ostream& out, const SString<X>& string);
还有一个值得吐槽的地方,你的+
运算符太迷了,改为+=
或者像+(const SString&)
一样不要返回引用。。
PHP中文网2017-04-17 13:47:08
template<typename T>
SString<T>& SString<T>::operator+(T x) {
List<T> *t1, *t2, *t3;
t2 = new List<T>();
t2->num = x;
t2->next = NULL;
if (head == NULL) {
head = t2;
}else{
for (t1 = head; t1 != NULL; t1 = t1->next) {
t3 = t1;
}
t3->next = t2;
}
return *this;
}
template<typename T>
SString<T>& SString<T>::operator+(T x) {
List<T> *t1, *t2, *t3;
t2 = new List<T>();
t2->num = x;
t2->next = NULL;
if (head == NULL) {
head = t2;
}
else{
for (t1 = head; t1->next != NULL; t1 = t1->next)
;
t3 = t1;
t3->next = t2;
}
return *this;
}
改成这样vs里就能编译通过,原因我也不清楚,提示t3可能未初始化,但看逻辑循环体内一定会执行到
但我改后还遇到个问题
SString() {
head = NULL;
}
SString(const SString& string) {
List<T> *t;
for (t = string.head; t != NULL; t = t->next) {
*this + t->num;
}
}
你这的拷贝构造函数没初始化head,但是
template<typename T>
SString<T> SString<T>::operator+(const SString<T>& string) const
{
List<T> *t;
SString<T> s(*this);
for (t = string.head; t != NULL; t = t->next) {
cout << t->num;
s.operator+(t->num);
}
return s;
}
这里返回s的时候会调用拷贝构造函数,由于head未初始化,导致
*this + t->num;
这句话中传入是错误的head地址
但我不清楚为什么除了返回值的时候,其余时候调用拷贝构造函数都会将head初始化为NULL,新手还望解答。