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林創建綴問題解算器，它在最後的崩潰而循環到完成最後部分的方程。綴方程解算器C++ while循環堆
我稱之爲主最後while循環來解決什麼留在堆棧上，它掛在那兒，如果我從棧中彈出最後一個元素，將退出循環，並返回錯誤的答案。
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#include <iostream>
#include<stack>
#include<string>
#include <ctype.h>
#include <stdlib.h>
#include <stdio.h>
#include <sstream>
using namespace std;
#define size 30
int count=0;
int count2=0;
int total=0;
stack<string> prob;
char equ[size];
char temp[10];
string oper;
string k;
char t[10];
int j=0;
char y;
int solve(int f,int s, char o)
{
cout<<"f="<<f<<endl;
cout<<"s="<<s<<endl;
cout<<"o="<<o<<endl;
int a;
if (o== '*')//checks the operand stack for operator
{
cout << f << "*" << s << endl;
a= f*s;
}
if (o == '/')//checks the operand stack for operator
{
cout << f << "/" << s << endl;
if(s==0)
{
cout<<"Cant divide by 0"<<endl;
}
else
a= f/s;
}
if (o == '+')//checks the operand stack for operator
{
cout << f << "+" << s << endl;
a= f+s;
}
if (o == '-')//checks the operand stack for operator
{
cout << f << "-" << s << endl;
a= f-s;
}
return a;
}
int covnum()
{
int l,c;
k=prob.top();
for(int i=0;k[i]!='\n';i++)t[i]=k[i];
return l=atoi(t);
}
char covchar()
{
k=prob.top();
for(int i=0;k[i]!='\n';i++)t[i]=k[i];
return t[0];
}
void tostring(int a)
{
stringstream out;
out << a;
oper = out.str();
}
void charstack(char op)
{
oper=op;
prob.push(oper);
}
void numstack(char n[])
{
oper=n;
prob.push(oper);
}
void setprob()
{
int f,s;
char o;
char t;
int a;
int i;
t=covchar();
if(ispunct(t))
{
if(t=='(')
{
prob.pop();
}
if(t==')')
{
prob.pop();
}
else if(t=='+'||'-')
{
y=t;
prob.pop();
}
else if(t=='/'||'*')
{
y=t;
prob.pop();
}
}
cout<<"y="<<y<<endl;
i=covnum();
cout<<"i="<<i<<endl;
s=i;
prob.pop();
t=covchar();
cout<<"t="<<t<<endl;
if(ispunct(t))
{
o=t;
prob.pop();
}
i=covnum();
cout<<"i="<<i<<endl;
f=i;
prob.pop();
t=covchar();
if (t=='('||')')
{
prob.pop();
}
a=solve(f,s, o);
tostring(a);
prob.push(oper);
cout<<"A="<<prob.top()<<endl;
}
void postfix()
{
int a=0;
char k;
for(int i=0;equ[i]!='\0';i++)
{
if(isdigit(equ[i]))//checks array for number
{
temp[count]=equ[i];
count++;
}
if(ispunct(equ[i]))//checks array for operator
{
if(count>0)//if the int input is done convert it to a string and push to stack
{
numstack(temp);
count=0;//resets the counter
}
if(equ[i]==')')//if char equals the ')' then set up and solve that bracket
{
setprob();
i++;//pushes i to the next thing in the array
total++;
}
while(equ[i]==')')//if char equals the ')' then set up and solve that bracket
{
i++;
}
if(isdigit(equ[i]))//checks array for number
{
temp[count]=equ[i];
count++;
}
if(ispunct(equ[i]))
{
if(equ[i]==')')//if char equals the ')' then set up and solve that bracket
{
i++;
}
charstack(equ[i]);
}
if(isdigit(equ[i]))//checks array for number
{
temp[count]=equ[i];
count++;
}
}
}
}
int main()
{
int a=0;
char o;
int c=0;
cout<<"Enter Equation: ";
cin>>equ;
postfix();
while(!prob.empty())
{
setprob();
a=covnum();
cout<<a<<" <=="<<endl;
prob.pop();
cout<<prob.top()<<"<top before c"<<endl;
c=covnum();
a=solve(c,a,y);
}
cout<<"Final Awnser"<<a<<endl;
system ("PAUSE");
return 0;
}
嚴肅的第一個建議：縮進並將代碼分解爲幾個函數......可能在您重寫代碼以便理解時，您也可以找到解決方案。 – paul23 2012-02-26 21:26:31
如上面請縮進我不能讀這個。 – 111111 2012-02-26 21:44:22
這是功課嗎？ – 2012-02-26 21:51:38