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Prefix Expression Evaluation

With a given Prefix Expression, we will show you how to evaluate a prefix expression using stack.
 

Algorithm to evaluate Prefix Expression:


In this algorithm, we will use stack to store operands. The step are as follows:
  • Get the Prefix Expression String
  • While the end of Pretfix string read from right to left
    • If the current character is operand
      • Push it into Stack
    • End If
    • If the current character is operator
      • Pop stack for first_operand
      • Pop stack for second_operand
      • Evaluate the expression (first_operand)(operator)(second_operand) and push the result into stack
    • End If
  • End While
  • Pop stack for result and return it
 


 

Function to evaluate Prefix Expression


double evaluate_prefix_exp(char *prestr){
 STACK stk;
 int i = strlen(prestr);
 double op1, op2;
 stk.top = -1;
 while(--i != -1){
  if(prestr[i] == ' '){ //escape space
   continue;
  }
  if(isdig(prestr[i])){
   push( &stk, (double)(prestr[i]-'0')/*char to int*/ );
  }
  else{
   op1 = pop(&stk); // first poped number will be used as 1st operand
   op2 = pop(&stk);
   push( &stk, operation( prestr[i], op1, op2 ) );
  }
 }
 return pop( &stk );
}
 

Program to evaluate Prefix Expression:


#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define STACK_SIZE 50

typedef struct{
 int top;
 double stack[STACK_SIZE];
} STACK;

double evaluate_prefix_exp(char *);
void push(STACK *, double );
double pop(STACK *);
int isdig(char);
double operation(char, double, double);

int main(){
 double answer;
 char *prefix_str = "++1*234"; //108
 answer = evaluate_prefix_exp(prefix_str);
 printf("The answer of prefix expression \"%s\" =   %.2f", prefix_str, answer);
 return 0;
}

double evaluate_prefix_exp(char *prestr){
 STACK stk;
 int i = strlen(prestr);
 double op1, op2;
 stk.top = -1;
 while(--i != -1){
  if(prestr[i] == ' '){ //escape space
   continue;
  }
  if(isdig(prestr[i])){
   push( &stk, (double)(prestr[i]-'0')/*char to int*/ );
  }
  else{
   op1 = pop(&stk); // first poped number will be used as 1st operand
   op2 = pop(&stk);
   push( &stk, operation( prestr[i], op1, op2 ) );
  }
 }
 return pop( &stk );
}

int isdig(char c){
 return ( (c >= '0') && (c <= '9') ); } void push(STACK *s, double num){ if(s->top == STACK_SIZE-1){
  printf("Stack Overflow\n");
 }
 else{
  s->stack[++s->top] = num;
 }
}

double pop(STACK *s){
 if(s->top == -1){
  printf("Stack Underflow\n");
  exit(2);
 }
 else{
  return s->stack[s->top--];
 }
}

double operation(char c, double op1, double op2){
 switch(c){
  case '+' : return op1+op2;
  case '-' : return op1-op2;
  case '*' : return op1*op2;
  case '/' : return op1/op2;
  case '^' : return pow(op1, op2);
  default  : printf("Invalid operator %c used", c);
      exit(1);
 }
}
 
The answer of prefix expression "++1*234" =   11.00
 

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