CS301 Data Structure fall 2010 second assignment final idea solution

[Xpert]

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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]

[SIZE="6"]Where's the solution?[/SIZE]

[IQ-test]

Solution can be found after few hours.

[shahidjee7]

Solution can be found after few hours.

bhai bhej du solution

[Xpert]

IDEA Solution of cs301 second assignment

Code:

#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.