Question

4A. Generate three address code for the following C segment: (consider array elements of 8 5M bytes), Draw the quadruple for the generated three address code. S n*n for (i-2; is; i+-+) ) (le) fort j-2 i: jn; jj+i) alil FALSE;) count++

Answer 1

In three segment code an array is defined as like for eg. take b=int a[i] and elements are of 4 bytes

so 3 segment code of above is

t1=&a \\ generates the address of array a where it stored according to datatype for eg. here it will generate address of 4 bytes etc.

t2 = 4 * i \\ tells that element in array are 4 bytes

t3=t1+t2 \\ defines the address of the element in array where that i^th element is stored

so to retreive the value from that address we will use pointer that is

t4 = *t3 \\ extracts the value stored at address t3

b = t4

so this way we use arrays

In our above problem the 3 address codes are

s=n*n \\ as given in above code

i=2 \\ i variable is initialised with 2 in loop

l1: if( i > s) goto l2 \\ it is for loop condition above if it is false then program terminates and jumps to l2

t1 = &a \\ extracts the starting address of array a

t2= 8*i \\ as it is written that each element is of array is 8 bytes

t3 = t1+t2   \\ provides the address of the ith element

t4=*t3 \\ extract the ith element value

if(t4==TRUE) goto l4 \\ condition that if t4 that is ith element contains some value and is not empty that is if element is true or false if if conditions fails it will not continue add just increment loop and starts procedure again  with incremented value of i

t11=count+1   

count=t11   

j=2*i \\ intialise the variable j

l2: if( j > n) goto l4 \\ if condition of fails go to first loop

t5=&a \\ address of array a

t6=8*j \\ 8 byte is the length of elements j

t7=t5+t6 \\ address of the element stored in array

t8=*t7 \\ extraction of value of element at t7 address

t8=FALSE \\ value of element set by user for next iteration

t9=j+i

j=t9

goto l2 // loop continue until j satisfies the condition

l4: t10=i+1

i=t10

goto l1

l2: \\ loop terminates here

Quadruples

Operator	Argument1	Argument2	Result
*	n	n	s
=	2		i
>	i	s
&	a		t1
*	8	i	t2
+	t2	t1	t3
*,=	t3		t4
==	t4	TRUE
+	count	1	t11
=	t11		count
*	2	i	j
>	j	n
&	a		t5
*	8	j	t6
+	t5	t6	t7
*,=	t7		t8
=	FALSE		t8
+	j	1	t9
+	j	1	t10
=	t10		i