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Shift Reduce Parser in Compiler
  • Difficulty Level : Easy
  • Last Updated : 02 Jun, 2021

Prerequisite – Parsing | Set 2 (Bottom Up or Shift Reduce Parsers)
Shift Reduce parser attempts for the construction of parse in a similar manner as done in bottom up parsing i.e. the parse tree is constructed from leaves(bottom) to the root(up). A more general form of shift reduce parser is LR parser. 
This parser requires some data structures i.e. 
 

  • A input buffer for storing the input string.
  • A stack for storing and accessing the production rules.

Basic Operations – 

  • Shift: This involves moving of symbols from input buffer onto the stack.
  • Reduce: If the handle appears on top of the stack then, its reduction by using appropriate production rule is done i.e. RHS of production rule is popped out of stack and LHS of production rule is pushed onto the stack.
  • Accept: If only start symbol is present in the stack and the input buffer is empty then, the parsing action is called accept. When accept action is obtained, it is means successful parsing is done.
  • Error: This is the situation in which the parser can neither perform shift action nor reduce action and not even accept action.

Example 1 – Consider the grammar 
        S –> S + S 
        S –> S * S 
        S –> id 
Perform Shift Reduce parsing for input string “id + id + id”. 
 
 

Example 2 – Consider the grammar 
        E –> 2E2 
        E –> 3E3 
        E –> 4 
Perform Shift Reduce parsing for input string “32423”. 
 
 



Example 3 – Consider the grammar
                         S –>  ( L ) | a        
                         L –>  L , S | S    
Perform Shift Reduce parsing for input string “( a , ( a , a ) ) “.  

                        Stack                                                   Input Buffer                                                   Parsing Action                              
 ( a , ( a , a ) ) $Shift
$ (  a , ( a , a ) ) $Shift
$ ( a , ( a , a ) ) $Reduce S → a
$ ( S , ( a , a ) ) $Reduce L → S
$ ( L , ( a , a ) ) $Shift
$ ( L , ( a , a ) ) $Shift
$ ( L , ( a , a ) ) $Shift
$ ( L , ( a , a ) ) $Reduce S → a
$ ( L , ( S , a ) ) $Reduce L → S
$ ( L , ( L , a ) ) $Shift
$ ( L , ( L , a ) ) $Shift
$ ( L , ( L , a ) ) $Reduce S → a
$ ( L , ( L , S ) ) ) $Reduce L →L , S
$ ( L , ( L ) ) $Shift
$ ( L , ( L ) ) $Reduce S → (L)
$ ( L , S ) $Reduce L → L , S
$ ( L ) $Shift
$ ( L ) $Reduce S → (L)
$ S $Accept

Following is the implementation in C-
 

C++




// Including Libraries
#include <bits/stdc++.h>
using namespace std;
   
// Global Variables
int z = 0, i = 0, j = 0, c = 0;
   
// Modify array size to increase 
// length of string to be parsed
char a[16], ac[20], stk[15], act[10]; 
   
// This Function will check whether
// the stack contain a production rule 
// which is to be Reduce.
// Rules can be E->2E2 , E->3E3 , E->4
void check()
{
    // Coping string to be printed as action
    strcpy(ac,"REDUCE TO E -> ");  
       
    // c=length of input string
    for(z = 0; z < c; z++) 
    {
        // checking for producing rule E->4
        if(stk[z] == '4'
        {
            printf("%s4", ac);
            stk[z] = 'E';
            stk[z + 1] = '\0';
               
            //pinting action
            printf("\n$%s\t%s$\t", stk, a); 
        }
    }
           
    for(z = 0; z < c - 2; z++)
    {
        // checking for another production
        if(stk[z] == '2' && stk[z + 1] == 'E' && 
                                stk[z + 2] == '2'
        {
            printf("%s2E2", ac);
            stk[z] = 'E';
            stk[z + 1] = '\0';
            stk[z + 2] = '\0';
            printf("\n$%s\t%s$\t", stk, a);
            i = i - 2;
        }
           
    }
           
    for(z = 0; z < c - 2; z++)
    {
        //checking for E->3E3
        if(stk[z] == '3' && stk[z + 1] == 'E' && 
                                stk[z + 2] == '3'
        {
            printf("%s3E3", ac);
            stk[z]='E';
            stk[z + 1]='\0';
            stk[z + 1]='\0';
            printf("\n$%s\t%s$\t", stk, a);
            i = i - 2;
        }
    }
    return ; // return to main
}
   
// Driver Function
int main()
{
    printf("GRAMMAR is -\nE->2E2 \nE->3E3 \nE->4\n");    
       
    // a is input string
    strcpy(a,"32423"); 
       
    // strlen(a) will return the length of a to c
    c=strlen(a); 
       
    // "SHIFT" is copied to act to be printed
    strcpy(act,"SHIFT"); 
       
    // This will print Lables (column name)
    printf("\nstack \t input \t action"); 
       
    // This will print the initial 
    // values of stack and input
    printf("\n$\t%s$\t", a); 
       
    // This will Run upto length of input string
    for(i = 0; j < c; i++, j++) 
    {
        // Printing action
        printf("%s", act); 
           
        // Pushing into stack
        stk[i] = a[j];     
        stk[i + 1] = '\0';
           
        // Moving the pointer
        a[j]=' ';
           
        // Printing action
        printf("\n$%s\t%s$\t", stk, a); 
           
        // Call check function ..which will 
        // check the stack whether its contain
        // any production or not
        check(); 
    }
       
    // Rechecking last time if contain
    // any valid production then it will
    // replace otherwise invalid
    check(); 
       
    // if top of the stack is E(starting symbol)
    // then it will accept the input
    if(stk[0] == 'E' && stk[1] == '\0'
        printf("Accept\n");
    else //else reject
        printf("Reject\n");
}
// This code is contributed by Shubhamsingh10

C




//Including Libraries
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
 
//Global Variables
int z = 0, i = 0, j = 0, c = 0;
 
// Modify array size to increase
// length of string to be parsed
char a[16], ac[20], stk[15], act[10];
 
// This Function will check whether
// the stack contain a production rule
// which is to be Reduce.
// Rules can be E->2E2 , E->3E3 , E->4
void check()
{
    // Coping string to be printed as action
    strcpy(ac,"REDUCE TO E -> "); 
     
    // c=length of input string
    for(z = 0; z < c; z++)
    {
        //checking for producing rule E->4
        if(stk[z] == '4')
        {
            printf("%s4", ac);
            stk[z] = 'E';
            stk[z + 1] = '\0';
             
            //pinting action
            printf("\n$%s\t%s$\t", stk, a);
        }
    }
         
    for(z = 0; z < c - 2; z++)
    {
        //checking for another production
        if(stk[z] == '2' && stk[z + 1] == 'E' &&
                                stk[z + 2] == '2')
        {
            printf("%s2E2", ac);
            stk[z] = 'E';
            stk[z + 1] = '\0';
            stk[z + 2] = '\0';
            printf("\n$%s\t%s$\t", stk, a);
            i = i - 2;
        }
         
    }
         
    for(z=0; z<c-2; z++)
    {
        //checking for E->3E3
        if(stk[z] == '3' && stk[z + 1] == 'E' &&
                                stk[z + 2] == '3')
        {
            printf("%s3E3", ac);
            stk[z]='E';
            stk[z + 1]='\0';
            stk[z + 1]='\0';
            printf("\n$%s\t%s$\t", stk, a);
            i = i - 2;
        }
    }
    return ; //return to main
}
 
//Driver Function
int main()
{
    printf("GRAMMAR is -\nE->2E2 \nE->3E3 \nE->4\n");   
     
    // a is input string
    strcpy(a,"32423");
     
    // strlen(a) will return the length of a to c
    c=strlen(a);
     
    // "SHIFT" is copied to act to be printed
    strcpy(act,"SHIFT");
     
    // This will print Lables (column name)
    printf("\nstack \t input \t action");
     
    // This will print the initial
    // values of stack and input
    printf("\n$\t%s$\t", a);
     
    // This will Run upto length of input string
    for(i = 0; j < c; i++, j++)
    {
        // Printing action
        printf("%s", act);
         
        // Pushing into stack
        stk[i] = a[j];    
        stk[i + 1] = '\0';
         
        // Moving the pointer
        a[j]=' ';
         
        // Printing action
        printf("\n$%s\t%s$\t", stk, a);
         
        // Call check function ..which will
        // check the stack whether its contain
        // any production or not
        check();
    }
     
    // Rechecking last time if contain
    // any valid production then it will
    // replace otherwise invalid
    check();
     
    // if top of the stack is E(starting symbol)
    // then it will accept the input
    if(stk[0] == 'E' && stk[1] == '\0')
        printf("Accept\n");
    else //else reject
        printf("Reject\n");
}
// This code is contributed by Ritesh Aggarwal

Output –

GRAMMAR is -
E->2E2 
E->3E3 
E->4

stack      input      action
$    32423$    SHIFT
$3     2423$    SHIFT
$32      423$    SHIFT
$324       23$    REDUCE TO E -> 4
$32E       23$    SHIFT
$32E2        3$    REDUCE TO E -> 2E2
$3E        3$    SHIFT
$3E3         $    REDUCE TO E -> 3E3
$E         $    Accept

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