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 the shift-reduce parser is the LR parser.
This parser requires some data structures i.e.
- An input buffer for storing the input string.
- A stack for storing and accessing the production rules.
Basic Operations –
- Shift: This involves moving symbols from the 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 a production rule is popped out of a stack and LHS of a production rule is pushed onto the stack.
- Accept: If only the start symbol is present in the stack and the input buffer is empty then, the parsing action is called accept. When accepted 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-
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()
{ // Copying 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';
//printing 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 + 2]='\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 Labels (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 Variablesint z = 0, i = 0, j = 0, c = 0;
// Modify array size to increase // length of string to be parsedchar 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->4void check()
{ // Copying 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';
//printing 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 Functionint 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 Labels (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
Advantages:
- Shift-reduce parsing is efficient and can handle a wide range of context-free grammars.
- It can parse a large variety of programming languages and is widely used in practice.
- It is capable of handling both left- and right-recursive grammars, which can be important in parsing certain programming languages.
- The parse table generated for shift-reduce parsing is typically small, which makes the parser efficient in terms of memory usage.
Disadvantages:
- Shift-reduce parsing has a limited lookahead, which means that it may miss some syntax errors that require a larger lookahead.
- It may also generate false-positive shift-reduce conflicts, which can require additional manual intervention to resolve.
- Shift-reduce parsers may have difficulty in parsing ambiguous grammars, where there are multiple possible parse trees for a given input sequence.
- In some cases, the parse tree generated by shift-reduce parsing may be more complex than other parsing techniques.
Recommended Articles