Compiler Design Test 5

Question 1
1 / -0

Which of the following statement about peephole optimization is not correct?

A
It is done by examining a sliding window of target instructions and replacing sequences within the peephole by shorter sequence possible.

B
Code in the peephole must be contiguous

C
It helps in strength reduction.

D
It eliminates the unreachable code.

Question 2
1 / -0

Consider the following C-code.
int i = 0;
while( i < 1000)
{
printf(“%d”, i);
i++;
}
Which of the following optimization technique can be applied?

A
optimization not possible.

B
loop jamming

C
loop unrolling

D
code movement

Question 3
1 / -0

Which of the following statement about live variable analysis is correct?

A
If value of variable x is at point p could be used along some path in the flow graph starting at p then x is dead at p.

B
It helps in register allocation for basic blocks.

C
After a value is computed in a register, then it is necessary to store that value if it is dead at the end of the block.

D
If all registers are full, then we should use a register with a live value since that value need to be stored.

Question 4
1 / -0

Consider the code segment
int FUN()
{
int i, j, x, y, m, n;
n=20;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
if (i % 2)
{
x += ((4*j) + 5*i);
y += (7 + 4*j);
}
}
}
return x+y;
}
Which one of the following is/are true?

A
The code contains loop invariant computation

B
There is scope of dead code elimination in this code

C
There is scope of strength reduction in this code

D
There is scope of common sub-expression elimination in this code

Question 5
1 / -0

Consider the following set of statements:
S1: A DAG representation of a basic block is a directed a cyclic graph in which the nodes of the DAG represent the statements within the block and each child of a node corresponds to the statement that is the last definition of an operand used in the statement.
S2: A basic block is a maximal sequence of consecutive three-address statements in which flow of control can only enter at the first statement of the block and leave at the last statement without halting or branching except possibly at the last statement in the basic block.
S3: Register assignment is the process of deciding which IR values to keep in registers.
Which of the above statements is/are true?

A
Only S1

B
Both S1 and S2

C
Only S3

D
Both S1 and S3

Question 6
1 / -0

Which of the following statement is/are related with the code motion?

A
It decreases the amount of code in the loop.

B
It is a machine independent optimization

C
It consists of statements which can be moved outside the body of a loop without affecting the semantics of the program.

D
if the operands are known at compile type, then it performs the operation statically

Question 7
1 / -0

What is the result of following intermediate code after eliminating local common subexpression?
Given code :
t6 = 4 * i
x = a[t6]
t7 = 4 * i
t8 = 4 * j
t9 = a[t8]
a[t7] = t9
t10 = 4 * j
a[t10] = x
go to L

A
t6 = 4* i
x = a[t6]
t8 = 4 *j
t9 = a[t8]
a[t6] = t9
a[t8] = x
Go to L

B
t6 = 4* i
x = a[t7]
t8 = 4 *j
t9 = a[t8]
a[t7] = t9
a[t8] = x
Go to L

C
t6 = 4* i
x = a[t6]
t7 = 4 *j
t9 = a[t7]
a[t7] = t9
a[t9] = x
Go to L

D
None of these

Submit Test

Compiler Design...

Question 1

It is done by examining a sliding window of target instructions and replacing sequences within the peephole by shorter sequence possible.

Code in the peephole must be contiguous

It helps in strength reduction.

It eliminates the unreachable code.

Question 2

optimization not possible.

loop jamming

loop unrolling

code movement

Question 3

If value of variable x is at point p could be used along some path in the flow graph starting at p then x is dead at p.

It helps in register allocation for basic blocks.

After a value is computed in a register, then it is necessary to store that value if it is dead at the end of the block.

If all registers are full, then we should use a register with a live value since that value need to be stored.

Question 4

The code contains loop invariant computation

There is scope of dead code elimination in this code

There is scope of strength reduction in this code

There is scope of common sub-expression elimination in this code

Question 5

Only S1

Both S1 and S2

Only S3

Both S1 and S3

Question 6

It decreases the amount of code in the loop.

It is a machine independent optimization

It consists of statements which can be moved outside the body of a loop without affecting the semantics of the program.

if the operands are known at compile type, then it performs the operation statically

Question 7

t6 = 4* i
x = a[t6]
t8 = 4 *j
t9 = a[t8]
a[t6] = t9
a[t8] = x
Go to L

t6 = 4* i
x = a[t7]
t8 = 4 *j
t9 = a[t8]
a[t7] = t9
a[t8] = x
Go to L

t6 = 4* i
x = a[t6]
t7 = 4 *j
t9 = a[t7]
a[t7] = t9
a[t9] = x
Go to L

None of these

User

Question Analysis

Compiler Design...

Question 1

It is done by examining a sliding window of target instructions and replacing sequences within the peephole by shorter sequence possible.

Code in the peephole must be contiguous

It helps in strength reduction.

It eliminates the unreachable code.

Question 2

optimization not possible.

loop jamming

loop unrolling

code movement

Question 3

If value of variable x is at point p could be used along some path in the flow graph starting at p then x is dead at p.

It helps in register allocation for basic blocks.

After a value is computed in a register, then it is necessary to store that value if it is dead at the end of the block.

If all registers are full, then we should use a register with a live value since that value need to be stored.

Question 4

The code contains loop invariant computation

There is scope of dead code elimination in this code

There is scope of strength reduction in this code

There is scope of common sub-expression elimination in this code

Question 5

Only S1

Both S1 and S2

Only S3

Both S1 and S3

Question 6

It decreases the amount of code in the loop.

It is a machine independent optimization

It consists of statements which can be moved outside the body of a loop without affecting the semantics of the program.

if the operands are known at compile type, then it performs the operation statically

Question 7

t6 = 4* ix = a[t6]t8 = 4 *jt9 = a[t8]a[t6] = t9a[t8] = xGo to L

t6 = 4* ix = a[t7]t8 = 4 *jt9 = a[t8]a[t7] = t9a[t8] = xGo to L

t6 = 4* ix = a[t6]t7 = 4 *jt9 = a[t7]a[t7] = t9a[t9] = xGo to L

None of these

User

Question Analysis

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t6 = 4* i
x = a[t6]
t8 = 4 *j
t9 = a[t8]
a[t6] = t9
a[t8] = x
Go to L

t6 = 4* i
x = a[t7]
t8 = 4 *j
t9 = a[t8]
a[t7] = t9
a[t8] = x
Go to L

t6 = 4* i
x = a[t6]
t7 = 4 *j
t9 = a[t7]
a[t7] = t9
a[t9] = x
Go to L