postfix evaluation algorithm - c++

Question

here is my attempt for evaluation postfix evaluation
#include<iostream>
#include<string>
using namespace std;
template<class T>
class Stack
{
private:
T *s;int N;
public:
Stack(int maxn)
{
s=new T[maxn];
N=0;
}
int empty()const
{
return N==0;
}
void push(T k)
{
s[N++]=k;
}
T pop()
{
return s[--N];
}
};
int main()
{
//postfix evaluation
char *a="3+4*5";
int N=strlen(a);
Stack<int>save(N);
for(int i=0;i<N;i++)
{
if(a[i]=='+')
save.push(save.pop()+save.pop());
if(a[i]=='*')
save.push(save.pop()*save.pop());
if((a[i]>='0' && a[i]<='9'))
save.push(0);
while((a[i]>='0' && a[i]<='9'))
save.push(10*save.pop()+(a[i++]-'0'));
}
cout<<save.pop()<<" "<<endl;
return 0;
}
but instead of answer 23 because 4*5+3=23,it gives me answer 5,as i understood,this code gives me this result because,first it checks if there is + mark for i=0,which is not,then it checks if if it is *,this is also not,so it first push 0,then it evaluates 10*0+'3'-'0',which is equal to 3,(it would be pushed in stack),for i=1,a[i] is equal to 3,so it prints 3+, second pop is undefined,so i think it is error,please help me to fix it

Answer 1

This works with a little fix:
#include <iostream>
#include <cstring>
using namespace std;
template<class T>
class Stack
{
private:
T *s;
int N;
public:
Stack(int maxn)
{
s = new T[maxn];
N = 0;
}
int empty()const
{
return N == 0;
}
void push(T k)
{
s[N++] = k;
}
T pop()
{
return s[--N];
}
};
int main()
{
//postfix evaluation
const char *a = "3 4 5*+";
int N = strlen(a);
Stack<int>save(N);
for (int i = 0; i < N; i++)
{
if (a[i]=='+')
save.push(save.pop() + save.pop());
if (a[i]=='*')
save.push(save.pop() * save.pop());
if (a[i] >= '0' && a[i] <= '9')
{
save.push(0);
while (a[i] >= '0' && a[i] <= '9')
save.push(10 * save.pop() + a[i++] - '0');
i--;
}
}
cout << save.pop() << " " << endl;
return 0;
}
Output (ideone):
23
Now, if you remove that i--; which I added, the code will be skipping characters in a[] because of 2 increments of i, in a[i++] and for (int i = 0; i < N; i++).
Without i--; the output is (ideone):
9