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GATE | GATE CS 2010 | Question 37

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  • Last Updated : 03 Nov, 2022
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The program below uses six temporary variables a, b, c, d, e, f. 
 

 
    a = 1
    b = 10
    c = 20
    d = a+b
    e = c+d
    f = c+e
    b = c+e
    e = b+f
    d = 5+e
    return d+f

Assuming that all operations take their operands from registers, what is the minimum number of registers needed to execute this program without spilling?
 

(A)

2
 

(B)

3
 

(C)

4
 

(D)

6
 


Answer: (B)

Explanation:

All of the given expressions use at-most 3 variables, so we never need more than 3 registers. 

See  http://en.wikipedia.org/wiki/Register_allocation 

It requires minimum of 3 registers. 

Principle of Register Allocation : If a variable needs to be allocated to a register, the system checks for any free register available, if it finds one, it allocates. If there is no free register, then it checks for a register that contains a dead variable ( a variable whose value is not going to be used in the future ), and if it finds one then it allocates. Otherwise, it goes for Spilling ( it checks for a register whose value is needed after the longest time, saves its value into the memory, and then use that register for current allocation, later when the old value of the register is needed, the system gets it from the memory where it was saved and allocate it in any register which is available ). 

But here we should not apply spilling as directed in the question. 

Let\’s allocate the registers for the variables. 

a = 1 ( let\’s say register R1 is allocated for variable \’a\’ ) 

  

b = 10 ( R2 for \’b\’ , because value of \’a\’ is going to be used in the future, hence can not replace variable of \’a\’ by that of \’b\’ in R1) 

  

c = 20 ( R3 for \’c\’, because values of \’a\’ and \’b\’ are going to be used in the future, hence can not replace variable \’a\’ or \’b\’ by \’c\’ in R1 or R2 respectively) 

  

d = a+b ( now, \’d\’ can be assigned to R1 because R1 contains a dead variable which is \’a\’ and it is so-called because it is not going to be used in future, i.e. no subsequent expression uses the value of variable \’a\’) 

  

e = c+d ( \’e\’ can be assigned to R1, because currently R1 contains value of variable \’d\’ which is not going to be used in the subsequent expression.) 

Note: an already calculated value of a variable is used only by READ operation ( not WRITE), hence we have to see only on the RHS side of the subsequent expressions whether the variable is going to be used or not. 

  

f = c+e ( \’ f \’ can be assigned to R2, because vaule of \’b\’ in register R2 is not going to be used in subsequent expressions, hence R2 can be used to allocate for \’ f \’ replacing \’b\’ ) 

  

b = c+e ( \’ b \’ can be assigned to R3, because value of \’c\’ in R3 is not being used later ) 

  

e = b+f ( here \’e\’ is already in R1, so no allocation here, direct assignment ) 

  

d = 5+e ( \’d\’ can be assigned to either R1 or R3, because values in both are not used further, let\’s assign in R1 ) 

  

return d+f ( no allocation here, simply contents of registers R1 and R2 are added and returned) 

  

hence we need only 3 registers, R1 R2 and R3. 

 
 


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