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You can convert from postfix to infix through the use of stacks. Consider the following expression conversion:
54+67*+ -> ((5+4)+(6*7))
The way this can be achieved is that whenever you encounter an operator, pop the last two expressions and join them using the operator. Remember to include the open braces before the first expression and a close braces after the second expression. Check the given link below for the program:
What else can I help you with?
C plus plus program using a stacks converting a postfix-infix?
Yes
Why do compilers convert infix expressions to postfix?
people almost exclusively use infix notation to write mathematical expressions, computer languages almost exclusively allow programmers to use infix notation. However, if a compiler allowed infix expressions into the binary code used in the compiled version of a program, the resulting code would be larger than needed and very inefficient. Because of this, compilers convert infix expressions into postfix notation expressions, which have a much simpler set of rules for expression evaluation. Postfix notation gets its name from the fact that operators in a postfix expression follow the operands that they specify an operation on. Here are some examples of equivalent infix and postfix expressions Infix Notation Postfix Notation 2 + 3 2 3 + 2 + 3 * 6 3 6 * 2 + (2 + 3) * 6 2 3 + 6 * A / (B * C) + D * E - A - C A B C * / D E * + A C * - Where as infix notation expressions need a long list or rules for evaluation, postfix expressions need very few.
Who invented postfix and infix?
infix: old Egyptians/Assirs some thousands year before prefix: Jan Łukasiewicz (Polish Notation) postfix: Burks, Warren, and Wright (Reverse Polish Notation)
Convert infix to prefix to postfix?
(a + b) * c / ((x - y) * z)
Why you need convert a expression into postfix expression?
You convert an (infix) expression into a postfix expression as part of the process of generating code to evaluate that expression.
Prefix to postfix conversion using C programming?
#include<stdio.h> #include<conio.h> #include<string.h> char symbol,s[10]; int F(symbol) { switch(symbol) { case '+': case '-':return 2; case '*': case '/':return 4; case '^': case '$':return 5; case '(':return 0; case '#':return -1; default :return 8; } } int G(symbol) { switch(symbol) { case '+': case '-':return 1; case '*': case '/':return 3; case '^': case '$':return 6; case '(':return 9; case ')':return 0; default: return 7; } } void infix_to_postfix(char infix[],char postfix[]) { int top=-1,j=0,i,symbol; s[++top]='#'; for(i=0;i<strlen(infix);i++) { symbol=infix[i]; while(F(s[top])>G(symbol)) { postfix[j]=s[top--]; j++; } if(F(s[top])!=G(symbol)) s[++top]=symbol; else top--; } while(s[top]!='#') { postfix[j++]=s[top--]; } postfix[j]='\0'; } void main() { char infix[30],postfix[30]; clrscr(); printf("Enter the valid infix expression\n"); scanf("%s",infix); infix_to_postfix(infix, postfix); printf("postfix expression is \n %s", postfix); getch(); }
Which data structure is needed to convert infix notations to post fix notations?
stack is the basic data structure needed to convert infix notation to postfix
Example program of how to convert infix notation to postfix notation and prefix notation?
/**************************//**********cReDo**********//*****mchinmay@live.com***///C PROGRAM TO CONVERT GIVEN VALID INFIX EXPRESSION INTO POSTFIX EXPRESSION USING STACKS.#include#include#include#define MAX 20char stack[MAX];int top=-1;char pop();void push(char item);int prcd(char symbol){switch(symbol){case '+':case '-':return 2;break;case '*':case '/':return 4;break;case '^':case '$':return 6;break;case '(':case ')':case '#':return 1;break;}}int isoperator(char symbol){switch(symbol){case '+':case '-':case '*':case '/':case '^':case '$':case '(':case ')':return 1;break;default:return 0;}}void convertip(char infix[],char postfix[]){int i,symbol,j=0;stack[++top]='#';for(i=0;iprcd(stack[top]))push(symbol);else{while(prcd(symbol)
What is a c plus plus program that accepts a mathematical expression from a user and converts it to postfix expression and evaluates the result?
#include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> #define size 400 using namespace std; char infix[size]="\0",postfix[size]="\0",Stack[size]; int top; int precedence(char ch) { switch(ch) { case '^':return 5; case '/':return 4; case '*':return 4; case '+':return 3; case '-':return 3; default:return 0; } } char Pop() { char ret; if(top!=-1) { ret=Stack[top]; top--; return ret; } else return '#'; } char Topelem() { char ch; if(top!=-1) ch=Stack[top]; else ch='#'; return ch; } void Push(char ch) { if(top!=size-1) { top++; Stack[top]=ch; } } int braces(char* s) { int l,r; l=0;r=0; for(int i=0;s[i];i++) { if(s[i]=='(') l++; if(s[i]==')') r++; } if(l==r) return 0; else if(l<r) return 1; else return -1; } int main() { char ele,elem,st[2]; int T,prep,pre,popped,j=0,chk=0; cin>>T; while(T--) { j=0;chk=0;top=-1; strcpy(postfix," "); cin>>infix; chk=braces(infix); if(chk==0) { for(int i=0;infix[i];i++) { if(infix[i]=='(') { elem=infix[i]; Push(elem); } else if(infix[i]==')') { while((popped=Pop())!='(') { postfix[j++]=popped; } } else if(infix[i]=='^'infix[i]=='/'infix[i]=='*'infix[i]=='+'infix[i]=='-') { elem=infix[i]; pre=precedence(elem); ele=Topelem(); prep=precedence(ele); if(pre>prep) Push(elem); else { while(prep>=pre) { if(ele=='#') break; popped=Pop(); ele=Topelem(); postfix[j++]=popped; prep=precedence(ele); } Push(elem); } } else { postfix[j++]=infix[i]; } } while((popped=Pop())!='#') postfix[j++]=popped; postfix[j]='\0'; cout<<postfix; } } }
Algorithm to convert postfix notation into infix notation?
/**************************//**********cReDo**********//*****mchinmay@live.com***///C PROGRAM TO CONVERT GIVEN VALID INFIX EXPRESSION INTO POSTFIX EXPRESSION USING STACKS.#include#include#include#define MAX 20char stack[MAX];int top=-1;char pop();void push(char item);int prcd(char symbol){switch(symbol){case '+':case '-':return 2;break;case '*':case '/':return 4;break;case '^':case '$':return 6;break;case '(':case ')':case '#':return 1;break;}}int isoperator(char symbol){switch(symbol){case '+':case '-':case '*':case '/':case '^':case '$':case '(':case ')':return 1;break;default:return 0;}}void convertip(char infix[],char postfix[]){int i,symbol,j=0;stack[++top]='#';for(i=0;iprcd(stack[top]))push(symbol);else{while(prcd(symbol)
What is prefix expression?
Example: prefix: * 2 + 3 4 infix: 2 * (3+4) postfix: 2 3 4 + *
How do you convert a prefix expression to postfix using recursion?
struct stack { char ele; struct stack *next; }; void push(int); int pop(); int precedence(char); struct stack *top = NULL; int main() { char infix[20], postfix[20]; int i=0,j=0; printf("ENTER INFIX EXPRESSION: "); gets(infix); while(infix[i]!='\0') { if(isalnum(infix[i])) postfix[j++]=infix[i]; else { if(top==NULL) push(infix[i]); else { while(top!=NULL && (precedence(top->ele)>=precedence(infix[i]))) postfix[j++]=pop(); push(infix[i]); } } ++i; } while(top!=NULL) postfix[j++]=pop(); postfix[j]='\0'; puts(postfix); getchar(); return 0; } int precedence(char x) { switch(x) { case '^': return 4; case '*': case '/': return 3; case '+': case '-': return 2; default: return 0; } } void push(int x) { int item; struct stack *tmp; if(top==NULL) { top=(struct stack *)malloc(sizeof(struct stack)); top->ele=x; top->next=NULL; } else { tmp=top; top->ele=x; top->next=tmp; } } int pop() { struct stack *tmp; int item; if(top==NULL) puts("EMPTY STACK"); else if(top->next==NULL) { tmp=top; item=top->ele; top=NULL; free(tmp); } else { tmp=top; item=top->ele; top=top->next; free(tmp); } return item; }
