Infix to Postfix Stack C++ Output - c++

Question

#include<iostream>
#include<cstring>
#include<string>
#include "linkedStack.h"
#include <fstream>
#include <iomanip>
#include "ArgumentManager.h"
using namespace std;
int getPrecedence(char ch) //this function will decide and return precedence for the operators and operand
{
switch (ch)
{
case '/':return 2;
break;
case '*': return 2;
break;
case '+': return 1;
break;
case '-': return 1;
break;
default: return 0;
}
}
string infixToPostfix(const string& expression) // this function will convert the infix expression to postfix by passing the infix expression string
{
int size = expression.size(); //size of infix string
char infix[expression.size()]; //converting string into array of chars
strncpy(infix, expression.c_str(), sizeof(infix));
infix[sizeof(infix)] = '\0'; //adding 0 for array ending
char postfix[strlen(infix)]; // create a char array with the size of the infix string length
linkedStack s;
int precedence;
int i = 0;
int k = 0;
char ch;
//iterate the infix expression
while (i < size)
{
ch = infix[i];
//push opening parenthesis to stack
if (ch == '(')
{
s.push(ch);
i++;
continue;
}
if (ch == ')')
{
while (!s.empty() && s.top() != '(') // if closing parenthesis is found pop of all the elements and append it to the postfix expression till we encounter an opening parenthesis
{
postfix[k++] = s.top();
s.pop();
}
if (!s.empty())
{
s.pop(); // pop the opening parenthesis
}
i++;
continue;
}
precedence = getPrecedence(ch);
if (precedence == 0)
{
postfix[k++] = ch; // if an operand is found add it to the postfix expression
}
else
{
if (s.empty())
{
s.push(ch); //push operator onto stack if the stack is empty
}
else
{
while (!s.empty() && s.top() != '(' && precedence <= getPrecedence(s.top())) // pop of all the operators from the stack to the postfix expression till we see an operator with a lower precedence than the current operator
{
postfix[k++] = s.top();
s.pop();
}
s.push(ch); // push the current operator to stack
}
}
i++;
}
while (!s.empty()) // pop all of the the remaining operators in the stack and add it to postfix
{
postfix[k++] = s.top();
s.pop();
}
postfix[k] = '\0'; // add null character to end of character array for c string
string postfixStr = postfix; //covert the final postfix char array to a string
return postfixStr; // returns the final postfix expression as a string
}
int main(int argc, char* argv[])
{
ArgumentManager am(argc, argv);
string infilename = am.get("A");
string outfilename = am.get("C");
string expression1;
ifstream infile;
infile.open(infilename);
if (infile.good())
{
getline(infile, expression1); //gets the infix string from file
}
infile.close(); //close the infile stream
string expression2 = "12 + 3 / 4";
string postfix1 = infixToPostfix(expression1);
string postfix2 = infixToPostfix(expression2);
cout << expression1 << endl;
cout << postfix1 << endl;
cout << expression2 << endl;
cout << postfix2 << endl;
return 0;
}
Hi, I'm having trouble with my output with my infix to post fix converter using stack. For instance in the case of "12 + 3 / 4" the output should be "12 3 4 / +" with the characters evenly spaced by one white space. However, it is adding unnecessary white spaces or none at all. For the test case above you can see the problem I'm running into in the attached pic or below. The first line is the infix the second is the post fix output. There are two cases here.Output Example
25 + ( 4 * 5 ) - 12
25 4 5 * + 12-
12 + 3 / 4
12 3 4/+

Answer 1

1) This code :
int size = expression.size(); //size of infix string
char infix[expression.size()]; //converting string into array of chars
strncpy(infix, expression.c_str(), sizeof(infix));
infix[sizeof(infix)] = '\0'; //adding 0 for array ending
could be replaced by :
char * infix = new char [expression.length()+1];
std::strcpy (infix, expression.c_str());
2) This code pop values without repush them into s after.
while (!s.empty() && s.top() != '(' && precedence <= getPrecedence(s.top())) // pop of all the operators from the stack to the postfix expression till we see an operator with a lower precedence than the current operator
{
postfix[k++] = s.top();
s.pop();
}

Answer 2

Thanks for all the advice. In the end I did a few things to achieve the desired output. First, I increased the postfix array. Next, in two certain spots in the loop, whenever it appended to the postfix array I would also add a space.
Here:
while (!s.empty() && s.top() != '(' && precedence <= getPrecedence(s.top())) // pop of all the operators from the stack to the postfix expression till we see an operator with a lower precedence than the current operator
{
postfix[k++] = s.top();
postfix[k++] = ' ';
s.pop();
}
And here:
while (!s.empty()) // pop all of the the remaining operators in the stack and add it to postfix
{
postfix[k++] = s.top();
postfix[k++] = ' ';
s.pop();
}
Lastly, I used a separate function to trim any extra spaces in the final string.
A little messy but gets the job done.