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in Compiler Design by Junior (875 points) | 41 views
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If the conflicts are resolved in ABOVE manner...

show the above manner? 

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https://gateoverflow.in/87037/gate2005-83b

 But i didn't get can u explain

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E -> id | E+E | E * E ===> + and * in same level ===> equal precedence

But doubt at associativity due to + has left side and right side E

then it is Bottom-Up parsers ===> Right Most Derivation ===> + and * have right associativity

1 Answer

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Given grammar :

E→num

EE+E / EE

 

  + x num $ E
I0     I2   I1
I1 I3 I4   ACCEPT  
I2 R3 R3   R3  
I3     I2   I5
I4     I2   I6
I5 I3, R1 I4, R1   R1  
I6 I3, R2 I4,R2   R2  

 

Note: Shift-reduce conflict: Yacc’s default action in the case of a shift-reduce conflict is to choose the shift action. So when there is a situation to choose b/w shift and reduce (eg (I6,+)) we will choose shift.

# In the below table E(i) is used just for identification purpose.

STACK INPUT ACTION OUTPUT
I0 3x2+1$ (I0,num) : Shift I2  
I03I2 x2+1$ (I2,x) : Reduce by E->num E(1).val=num.val=3
I0E(1)I1 x2+1$ (I1,x) : Shift I4  
I0E(1)I1xI4 2+1$ (I4,num): Shift I2  
I0E(1)I1xI42I2 +1$ (I2,+) : Reduce by E->num E(2).val=num.val=2
I0E(1)I1xI4E(2)I6 +1$ (I6,+): Shift I3  
I0E(1)I1xI4E(2)I6+I3 1$ (I3,num): Shift I2  
I0E(1)I1xI4E(2)I6+I31I2 $ (I2,$): Reduce by E->num E(3).val=num.val=1
I0E(1)I1xI4E(2)I6+I3E(3)I5 $ (I5,$): Reduce by E->E+E E(4).val=E(2).val+E(3).val=2+1=3
I0E(1)I1xI4E(4)I6 $ (I6,$): Reduce by E->ExE E(5).val=E(1).valxE(4).val=3x3=9
I0E(5)I1 $ (I1,$): ACCEPT  

 

by Boss (23.5k points)
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really need to do this?
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No but i wanted to give an elaboration..

The solution at https://gateoverflow.in/87037/gate2005-83b was not much clear to me..I understood the logic after building this table and all.. but the one Debashish Deka Sir added at https://gateoverflow.in/1405/gate2005-83a is also very detailed..

Deepalitrapti didn't get the solution in the first link..So i thought to give these details..

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ok...
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