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I need to implement a stack and this is what I have so far. After that I need to

ID: 3683424 • Letter: I

Question

I need to implement a stack and this is what I have so far.

After that I need to write a program called infixconverter that takes in a string containing an infix expression including (,), +, -, * and the integers and then outputs the postfix expression. I need to use the stack class for this as well.

  public Stack(int c)

   {

size = c;

stackArray = new long[size];

top = -1;

   }

public boolean isEmpty()

{

   return (top == -1);

}

   public long pop()

   {

       return stackArray[top--];

   }

   public long top()

   {

   return stackArray[top];

   }

   public void push(long j)

   {

   stackArray[top++] = j;

   }

   //You are welcome to add any other private functions.

}

Explanation / Answer

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

// Stack type
struct Stack
{
int top;
unsigned capacity;
int* array;
};

// Stack Operations
struct Stack* createStack( unsigned capacity )
{
struct Stack* stack = (struct Stack*) malloc(sizeof(struct Stack));

if (!stack)
return NULL;

stack->top = -1;
stack->capacity = capacity;

stack->array = (int*) malloc(stack->capacity * sizeof(int));

if (!stack->array)
return NULL;
return stack;
}
int isEmpty(struct Stack* stack)
{
return stack->top == -1 ;
}
char peek(struct Stack* stack)
{
return stack->array[stack->top];
}
char pop(struct Stack* stack)
{
if (!isEmpty(stack))
return stack->array[stack->top--] ;
return '$';
}
void push(struct Stack* stack, char op)
{
stack->array[++stack->top] = op;
}


// A utility function to check if the given character is operand
int isOperand(char ch)
{
return (ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z');
}

// A utility function to return precedence of a given operator
// Higher returned value means higher precedence
int Prec(char ch)
{
switch (ch)
{
case '+':
case '-':
return 1;

case '*':
case '/':
return 2;

case '^':
return 3;
}
return -1;
}


// The main function that converts given infix expression
// to postfix expression.
int infixToPostfix(char* exp)
{
int i, k;

// Create a stack of capacity equal to expression size
struct Stack* stack = createStack(strlen(exp));
if(!stack) // See if stack was created successfully
return -1 ;

for (i = 0, k = -1; exp[i]; ++i)
{
// If the scanned character is an operand, add it to output.
if (isOperand(exp[i]))
exp[++k] = exp[i];

// If the scanned character is an ‘(‘, push it to the stack.
else if (exp[i] == '(')
push(stack, exp[i]);
  
// If the scanned character is an ‘)’, pop and output from the stack
// until an ‘(‘ is encountered.
else if (exp[i] == ')')
{
while (!isEmpty(stack) && peek(stack) != '(')
exp[++k] = pop(stack);
if (!isEmpty(stack) && peek(stack) != '(')
return -1; // invalid expression
else
pop(stack);
}
else // an operator is encountered
{
while (!isEmpty(stack) && Prec(exp[i]) <= Prec(peek(stack)))
exp[++k] = pop(stack);
push(stack, exp[i]);
}

}

// pop all the operators from the stack
while (!isEmpty(stack))
exp[++k] = pop(stack );

exp[++k] = '';
printf( "%s ", exp );
}

// Driver program to test above functions
int main()
{
char exp[] = "a+b*(c^d-e)^(f+g*h)-i";
infixToPostfix(exp);
return 0;
}