Miscellaneous Problems in Compiler Design

Three Address Code | DAGs | Basic Blocks & Flow Graphs-

 

Before you go through this article, make sure that you have gone through the previous articles on-

• Three Address Code
• Basic Blocks and Flow Graphs
• Directed Acyclic Graphs

 

In this article, we will solve Miscellaneous Problems based on these Concepts.

 

PRACTICE PROBLEMS BASED ON ABOVE CONCEPTS-

 

Problem-01:

 

Generate three address code for the following code-

 

c = 0

do

{

if (a < b) then

x++;

else

x–;

c++;

} while (c < 5)

 

Solution-

 

Three address code for the given code is-

1. c = 0
2. if (a < b) goto (4)
3. goto (7)
4. T1 = x + 1
5. x = T1
6. goto (9)
7. T2 = x – 1
8. x = T2
9. T3 = c + 1
10. c = T3
11. if (c < 5) goto (2)

 

Problem-02:

 

Generate three address code for the following code-

 

while (A < C and B > D) do

if A = 1 then C = C + 1

else

while A <= D

do A = A + B

 

Solution-

 

Three address code for the given code is-

1. if (A < C) goto (3)
2. goto (15)
3. if (B > D) goto (5)
4. goto (15)
5. if (A = 1) goto (7)
6. goto (10)
7. T1 = c + 1
8. c = T1
9. goto (1)
10. if (A <= D) goto (12)
11. goto (1)
12. T2 = A + B
13. A = T2
14. goto (10)

 

Problem-03:

 

Generate three address code for the following code-

 

switch (ch)

{

case 1 : c = a + b;

break;

case 2 : c = a – b;

break;

}

 

Solution-

 

Three address code for the given code is-

 

if ch = 1 goto L1

if ch = 2 goto L2

L1:

T1 = a + b

c = T1

goto Last

L2:

T1 = a – b

c = T2

goto Last

Last:

 

Problem-04:

 

Construct a DAG for the following three address code-

1. a = b + c
2. t1 = a x a
3. b = t1 + a
4. c = t1 x b
5. t2 = c + b
6. a = t2 + t2

 

Solution-

 

Directed acyclic graph for the given three address code is-

 

 

Problem-05:

 

Consider the following code-

 

prod = 0 ;

i = 1 ;

do

{

prod = prod + a[ i ] x b[ i ] ;

i = i + 1 ;

} while (i <= 10) ;

 

1. Compute the three address code.
2. Compute the basic blocks and draw the flow graph.

 

Solution-

 

Part-01:

 

Three address code for the given code is-

 

prod = 0

i = 1

T1 = 4 x i

T2 = a[T1]

T3 = 4 x i

T4 = b[T3]

T5 = T2 x T4

T6 = T5 + prod

prod = T6

T7 = i + 1

i = T7

if (i <= 10) goto (3)

 

Part-02:

 

Step-01:

 

We identify the leader statements as-

• prod = 0 is a leader because first statement is a leader.
• T1 = 4 x i is a leader because target of conditional or unconditional goto is a leader.

 

Step-02:

 

The above generated three address code can be partitioned into 2 basic blocks as-

 

 

Step-03:

 

The flow graph is-

 

 

To gain better understanding about these Miscellaneous Problems,

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