【发布时间】:2015-09-12 17:09:58
【问题描述】:
我目前正在做一个项目,使用链表形式的堆栈从后缀转换为中缀。我目前正在尝试将整行作为字符串读取,然后将其放入字符数组中,然后当发现符号时,将一个元素放入右操作数中,将另一个元素放入左操作数中,然后将其打印回来,包括运算符。但是,在将第一项放入左侧操作数然后弹出堆栈后,我无法将另一项放入右侧操作数中。可能是什么问题呢?我用我的流行功能。
这是我的代码:
#include "stack.h"
stack::~stack()
{
cout<<"Inside !stack \n";
while(s_top != 0)
{
pop();
}
}
void stack::pop()
{
cout<<"Inside pop \n";
stack_node *p;
if (s_top != 0)
{
p = s_top;
s_top = s_top->next;
delete p;
}
}
void stack::push(char a)
{
cout<<"Inside push \n";
stack_node *p = new stack_node;
p->data = a;
p->next = s_top;
s_top = p;
}
void stack::print()
{
cout<<"Inside print \n";
for(stack_node *p = s_top; p!=0; p=p->next)
{
cout<<p->data<<endl;
}
}
stack_element stack::top()
{
cout<<"Inside top \n";
if (s_top == 0)
{
exit(1);
}
else
{
return s_top->data;
}
}
/*stack::stack(const stack & Org)
{
cout<<"Inside the Copy Constructor\n";
stack_node *p=Org.s_top;
(*this).s_top = 0;
while(p!=0)
{
(*this).push(p->data);
p=p->next;
}
}
这是我的cpp,它不能完全工作
#include "stack.h"
string convert(string expression){
stack c;
string post = " ";
string rightop="";
string leftop="";
string op =" ";
for (int i =0; i<expression.length();i++){
c.push(expression[i]);
if(expression[i]=='*'||'+'||'-'||'/'){
cout<<c.top()<<endl;
leftop=c.top();
c.pop();
rightop=c.top();
cout<<rightop<<endl;
c.pop();
op=c.top();
c.pop();
}
}
}
int main(){
string expression;
cout<<" Enter a Post Fix expression: ";
getline(cin,expression);
convert(expression);
return 0;
}
【问题讨论】:
标签: c++ linked-list stack postfix-notation infix-notation