CS301 Data Structure fall 2010 second assignment final idea solution

[Xpert]

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Instructions
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o Student ID is not mentioned in the assignment File or name of file is other than student ID.

Note: Kindly check the template file for submitting assignment. (see announcement)
Uploading instructions

Your Submission must include:

1. A working Make File (Dev-C++ project File).
2. All the Source Code (.h and .cpp files) necessary to compile and run your program.
3. Place all the files in a folder then Zip that folder and Upload it on VULMS

Note: Use Dev-C++ 4.9.9.2 which is available at VULMS (Download section) or you can use Dev-C++ 4.0 or any other version higher than 4.0.
Question:
Write a program in c++ that takes an infix expression from user through command line argument and convert it into a postfix expression. You are required to use the stack data structure for the implementation of this task.




Note: You can’t use built-in library function or templates for stack functions. All the source code should be written manually.

[Asim Ali]

Where's the solution?

[IQ-test]

Solution can be found after few hours.

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[shahidjee7]

Solution can be found after few hours.

bhai bhej du solution

[Xpert]

IDEA Solution of cs301 second assignment

#include<iostream.h>
#include<string.h>
#include<stdlib.h>
#include<ctype.h>
class expression
{
private:
 char infix[100];
 char stack[200];
 int top;
 int r;
 char postfix[100];
public:
 void convert();
 int input_p(char);
 int stack_p(char);
 int rank(char);
};
int expression::input_p(char c)
{
 if(c==’+’ || c==’-')
  return 1;
 else if(c==’*’ || c==’/')
  return 3;
 else if(c==’^')
  return 6;
 else if(isalpha(c)!=0)
  return 7;
 else if(c==’(‘)
  return 9;
 else if(c==’)')
  return 0;
 else
 {
  cout<<”Invalid expression ::input error\n”;
  exit(0);
 }
}
int expression::stack_p(char c)
{
 if(c==’+’ || c==’-')
  return 2;
 else if(c==’*’ || c==’/')
  return 4;
 else if(c==’^')
  return 5;
 else if(isalpha(c)!=0)
  return 8;
 else if(c==’(‘)
  return 0;
 else
 {
  cout<<”Invalid expression  ::stack error\n”;
  exit(0);
 }
}
int expression::rank(char c)
{
 if(c==’+’ || c==’-')
  return -1;
 else if(c==’*’ || c==’/')
  return -1;
 else if(c==’^')
  return -1;
 else if(isalpha(c)!=0)
  return 1;
 else
 {
  cout<<”Invalid expression ::in rank\n”;
  exit(0);
 }
}
void expression::convert()
{
 cout<<”\n*************************************************\n”
  <<”This program converts the given infix expression\n”
  <<”in to postfix form”
                <<”\n*************************************************\n”;
 cout<<”Enter an infix expression ::\n”;
 cin>>infix;
 int l=strlen(infix);
 infix[l]=’)';
 infix[l+1]=”;
 //Convertion starts
 top=1;
 stack[top]=’(‘;
 r=0;
 int x=-1;
 int i=0;
 char next=infix[i];
 while(next!=”)
 {
  //Pop all the elements to outputin stack which have higher precedence
  while( input_p(next) < stack_p(stack[top]) )
  {
   if(top<1)
   {
    cout<<”invalid expression ::stack error\n”;
    exit(0);
   }
   postfix[++x]=stack[top];
   top–;
   r=r+rank(postfix[x]);
   
   if(r<1)
   {
    cout<<”Invalid expression  ::r<1\n”;
    exit(0);
   }
  }
  if(input_p( next ) != stack_p( stack[top]))
   stack[++top]=next;
  else
   top–;
  i++;
  next=infix[i];
 }
 postfix[++x]=”;
 if(r!=1 || top!=0)
 {
  cout<<”Invalid expression ::error in rank or stack\n”;
  exit(0);
 }
 cout<<”\n\nThe corresponding postfix expression is ::\n”;
 cout<<postfix<<endl;
}
int main()
{
 expression obj;
 obj.convert();
 return 0;

[Xpert]

i guess this solution is good for students i have not found any comments.

[makyle]

there is no stake information and compile error

[rabeel]

it is just for idea dear.