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How do you convert infix to postfix without using data structures?

Anonymous

∙ 14y ago

Updated: 8/18/2019

Without data-structures you cannot even store expressions, let alone convert or evaluate them.

Wiki User

∙ 14y ago

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Q: How do you convert infix to postfix without using data structures?

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Related Questions

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.

Convert infix to prefix to postfix?

(a + b) * c / ((x - y) * z)

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

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.

Write an algorithm in c to convert an infix expression to a postfix expressionexecute your algorithm with the following infix expression as your input. m nk pgh a bc?

An algorithm can not be written with the following infix expression without knowing what the expression is. Once this information is included a person will be able to know how to write the algorithm.

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(); }

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)

C plus plus program using a stacks converting a postfix-infix?

Yes

Which data structure convert logical to physical address?

Linear data structure is used to convert the logical address to physical address .Stack is used in this and the various conversion such as postfix,prefix and infix notation are come in this

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)

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)

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; }

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