Stacks -Converting Postfix Expression to Infix Expresion C++

Question

Hi

Write a program write a program called "infix.cpp", that uses a stack to convert a postfix expression to the corresponding fully-parenthesized infix expression. Consider the following examples:

the postfix expression a b + c d - * will be converted to the infix ((a + b) * (c - d))
the the postfix expression a b + will be converted to the infix (a + b)
the postfix expression a b / c d / / will be converted to infix ((a / b) / (c / d))

Your program should ask the user for a postfix expression as input, and it should output the corresponding fully-parenthesized infix expression. The program should also ask the user if he/she would like to do another conversion. If so, the user should be able to enter another posfix expression; otherwise the program should terminate. Also, the stack must be implemented using a singly-linked list. Your driver, infix.cpp, should include the definition and declaration files for the class STACK, stack.cpp and stack.h, respectively.

**Using CodePad is helpful to understand your code.

Otherwise just submit your code neatly (understandable)

THANK YOU!

Vulpes · Answer

OK, try this: // stack.h #include using namespace std; struct node { string data; struct node *next; }; class stack { struct node *top; public: stack() { top = 0; } void push(string data); string pop(); }; // stack.cpp #include "stack.h" void stack::push(string data) { struct node *pn; pn = new node(); pn->data = data; pn->next = (top != 0) ? top : 0; top = pn; } string stack::pop() { struct node *pn; if (top == 0) return "\0" ; // slight alteration on this line pn = top; string result = pn->data; top = top->next; delete pn; return result; } // infix.cpp #include "stack.cpp" #include #include #include using namespace std; int main() // a lot of alterations in here { int error; stack s; string postfix, infix, operand1, operand2, expr, yn; do { cout getline(cin, postfix); error = 0; for(int i = 0; i { string c = postfix.substr(i, 1); if (c == "+" || c == "-" || c == "*" || c == "/") { operand2 = s.pop(); if (operand2 == "\0") { cout error = 1; break; } operand1 = s.pop(); if (operand1 == "\0") { cout error = 1; break; } expr = "(" + operand1 + " " + c + " " + operand2 + ")"; s.push(expr); } else if (c == " ") { continue; } else { s.push(c); } } if (!error) { infix = s.pop(); if (s.pop() == "\0") { cout } else { cout // make sure stack is clear while(s.pop() != "\0"); } } cout cin >> yn; getchar(); // swallow new line character cout } while(yn == "y" || yn == "Y"); return 0; }

Vulpes · Answer

Here's my own attempt: // stack.h #include using namespace std; struct node { string data; struct node *next; }; class stack { struct node *top; public: stack() { top = 0; } void push(string data); string pop(); }; // stack.cpp #include "stack.h" void stack::push(string data) { struct node *pn; pn = new node(); pn->data = data; pn->next = (top != 0) ? top : 0; top = pn; } string stack::pop() { struct node *pn; if (top == 0) return '\0'; pn = top; string result = pn->data; top = top->next; delete pn; return result; } // infix.cpp #include "stack.cpp" #include #include #include using namespace std; int main() { stack s; string postfix, infix, operand1, operand2, expr, yn; do { cout cin >> postfix; for(int i = 0; i { string c = postfix.substr(i, 1); if (c == "+" || c == "-" || c == "*" || c == "/") { operand2 = s.pop(); operand1 = s.pop(); expr = "(" + operand1 + c + operand2 + ")"; s.push(expr); } else { s.push(c); } } infix = s.pop(); cout cout cin >> yn; cout } while(yn == "y" || yn == "Y"); return 0; }

SUNIL GUTTA · Answer

Well thank you for all the help u given .. There has been a modification lately sorry for this . Can you provide the following PLEASE ? 4. for the postfix expression a b / c * + the program should print the error message "too many operators and not enough operand". 5. for the postfix expression a b c d / + e * f the program should print the error message "too many operands and not enough operators". NOTE ::: Include one space between operands ( eg. a b c d ) and operator (eg. + - * /) 2. The only operators to consider are +, -, * and / (If you can, please post the new and full code) Thank you so much friend Cheers & have a nice day

Vulpes · Answer

Yes, I always test complete programs before posting and it compiles and works fine (VS 2010). Here's some sample input and output: Enter postfix expression ab+cd-* The infix expression is ((a+b)*(c-d)) Convert another y/n : n It sounds like you've somehow defined (or included) the stack class twice. I'd check the #includes in all three files to see whether this is what you've done.

Ken H · Answer

hi SUNIL GUTTA , I once wrote a Reverse Polish algorithm using C#.It is the use of stack.

Jaganathan Bantheswaran · Answer

What is the your effort on this?

SUNIL GUTTA · Answer

TY mate u made my day :) will debug and learn how u did & any doubts i will come back to you .. cheers Ty very much .. Hey if u had time can u hint me the logic , its a different type of recursive problem in java .. i posted in java section Have a nice day

SUNIL GUTTA · Answer

Humm ty for such quick response :) i will check and do it :) Ty ur always the best cheers

SUNIL GUTTA · Answer

Hi i am facing fallowing errors Did you try this code in your compiler? For some reason there are a lot of errors. help it out dude 3)"public: class std::basic_string __thiscall stack::pop(void)" (?pop@stack@@QAE?AV?$basic_string@DU?$char_traits@D@std@@V?$allocator@D@2@@std@@XZ) already defined in infix.obj (4) fatal error LNK1169: one or more multiply defined symbols found 1) (1) warning C4018: ' cheers Ty

SUNIL GUTTA · Answer

Jaganathan Bantheswaran ty for concerned reply At first this kiran i.e sunil's friend / / Well developed the complete code :) everything is fine .. but my code looks bigger and i felt it may degrade performance :) . So came to u guys . thought of comparing ur code with mine and rectify my way of coding etc This forum is fantastic such genius minds so :) Happy very happy TY for reply Ken H H ave a nice day friends