Registers for holding values: The parameter “x” is been passed to function in register “%rdi”. Once loop is entered, there is no reference to register. The register “%rax” is initialized to “x”. The register “%rcx” is initialized to “x*x”...

Question

Want to see the full answer?

Answer 1

Expert Solution & Answer

Chapter 3.6, Problem 3.23PP

A.

Explanation of Solution

Registers for holding values:

The parameter “x” is been passed to function in register “%rdi”.

Once loop is entered, there is no reference to register.

The register “%rax” is initialized to “x”.

The register “%rcx” is initialized to “x*x”...

B.

Explanation of Solution

Compiler elimination of pointer variable:

The pointer “p” always points to x as determined by compiler.

The expression “(*p)++” would increment x...

C.

Explanation of Solution

Modified annotated code:

long dw_loop(long x)

x in %rdi

dw_loop:

movq %rdi, %rax;Copy x to %rax

movq %rdi, %rcx

imulq %rdi, %rcx;Compute y = xx

leaq (%rdi, %rdi), %rdx;Compute n = 2x

.L2:;Loop

leaq 1(%rcx, %rax), %rax;Compute x+=y+1

subq $1, %rdx;Decrement n

testq %rdx, %rdx;Test n

jg ...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

1. We wish to compare the performance of two different machines: M1 and M2. The following measurements have been made on these machines: Program Time on M1 Time on M2 1 10 seconds 5 seconds 2 3 seconds 4 seconds Which machine is faster for each program, and by how much? 2. For M1 and M2 of problem 1, the following additional measurements are made:. Find the instruction execution rate (instructions per second) for each machine when running program 1. Program Instructions executed on M1 Instructions executed on M2 1 200 x 106 160 x 106 3. For M1 and M2 of problem 1, if the clock rates are 200 MHz and 300 MHz, respectively, find the CPI for program 1 on both machines using the data provided in problems 1 and 2. 4. You are going to enhance a machine, and there are two possible improvements: either make multiply instructions run four times faster than before or make memory access instructions run two times faster than before. You…

A program has the following breakdown: 25% ld (50% of them directly followed by a dependent instruction),25% sd, 30% r_type, 20% beq (80% of them are taken. Branches are calculated in the third cycle. What is the average CPI of the program when run on the pipelined RISC V implementation in the textbook?

Here is my question from homework: int x; short y, z; cin>>x; y=x; z= y+2; cout<<"First number is: "<<x<<endl; cout<<"Second number is: "<<y<<endl; cout<<"Third number is: "<<z<<endl; End of HW problem My question is how do I utilize short variable type in MIPS assembly code and get it to work with int data type?

Answer 2

Expert Solution & Answer

Chapter 3.6, Problem 3.23PP

A.

Explanation of Solution

Registers for holding values:

The parameter “x” is been passed to function in register “%rdi”.

Once loop is entered, there is no reference to register.

The register “%rax” is initialized to “x”.

The register “%rcx” is initialized to “x*x”...

B.

Explanation of Solution

Compiler elimination of pointer variable:

The pointer “p” always points to x as determined by compiler.

The expression “(*p)++” would increment x...

C.

Explanation of Solution

Modified annotated code:

long dw_loop(long x)

x in %rdi

dw_loop:

movq %rdi, %rax;Copy x to %rax

movq %rdi, %rcx

imulq %rdi, %rcx;Compute y = xx

leaq (%rdi, %rdi), %rdx;Compute n = 2x

.L2:;Loop

leaq 1(%rcx, %rax), %rax;Compute x+=y+1

subq $1, %rdx;Decrement n

testq %rdx, %rdx;Test n

jg ...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

Chapter 3.6, Problem 3.23PP

A.

Explanation of Solution

Registers for holding values:

The parameter “x” is been passed to function in register “%rdi”.

Once loop is entered, there is no reference to register.

The register “%rax” is initialized to “x”.

The register “%rcx” is initialized to “x*x”...

B.

Explanation of Solution

Compiler elimination of pointer variable:

The pointer “p” always points to x as determined by compiler.

The expression “(*p)++” would increment x...

C.

Explanation of Solution

Modified annotated code:

long dw_loop(long x)

x in %rdi

dw_loop:

movq %rdi, %rax;Copy x to %rax

movq %rdi, %rcx

imulq %rdi, %rcx;Compute y = xx

leaq (%rdi, %rdi), %rdx;Compute n = 2x

.L2:;Loop

leaq 1(%rcx, %rax), %rax;Compute x+=y+1

subq $1, %rdx;Decrement n

testq %rdx, %rdx;Test n

jg ...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 3.6, Problem 3.23PP

A.

Explanation of Solution

Registers for holding values:

The parameter “x” is been passed to function in register “%rdi”.

Once loop is entered, there is no reference to register.

The register “%rax” is initialized to “x”.

The register “%rcx” is initialized to “x*x”...

B.

Explanation of Solution

Compiler elimination of pointer variable:

The pointer “p” always points to x as determined by compiler.

The expression “(*p)++” would increment x...

C.

Explanation of Solution

Modified annotated code:

long dw_loop(long x)

x in %rdi

dw_loop:

movq %rdi, %rax;Copy x to %rax

movq %rdi, %rcx

imulq %rdi, %rcx;Compute y = xx

leaq (%rdi, %rdi), %rdx;Compute n = 2x

.L2:;Loop

leaq 1(%rcx, %rax), %rax;Compute x+=y+1

subq $1, %rdx;Decrement n

testq %rdx, %rdx;Test n

jg ...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 3.6, Problem 3.23PP

A.

Explanation of Solution

Registers for holding values:

The parameter “x” is been passed to function in register “%rdi”.

Once loop is entered, there is no reference to register.

The register “%rax” is initialized to “x”.

The register “%rcx” is initialized to “x*x”...

B.

Explanation of Solution

Compiler elimination of pointer variable:

The pointer “p” always points to x as determined by compiler.

The expression “(*p)++” would increment x...

C.

Explanation of Solution

Modified annotated code:

long dw_loop(long x)

x in %rdi

dw_loop:

movq %rdi, %rax;Copy x to %rax

movq %rdi, %rcx

imulq %rdi, %rcx;Compute y = xx

leaq (%rdi, %rdi), %rdx;Compute n = 2x

.L2:;Loop

leaq 1(%rcx, %rax), %rax;Compute x+=y+1

subq $1, %rdx;Decrement n

testq %rdx, %rdx;Test n

jg ...

Answer 8

Chapter 3.6, Problem 3.23PP

A.

Explanation of Solution

Registers for holding values:

The parameter “x” is been passed to function in register “%rdi”.

Once loop is entered, there is no reference to register.

The register “%rax” is initialized to “x”.

The register “%rcx” is initialized to “x*x”...

Answer 9

Chapter 3.6, Problem 3.23PP

A.

Explanation of Solution

Registers for holding values:

The parameter “x” is been passed to function in register “%rdi”.

Once loop is entered, there is no reference to register.

The register “%rax” is initialized to “x”.

The register “%rcx” is initialized to “x*x”...

Answer 10

B.

Explanation of Solution

Compiler elimination of pointer variable:

The pointer “p” always points to x as determined by compiler.

The expression “(*p)++” would increment x...

Answer 11

B.

Explanation of Solution

Compiler elimination of pointer variable:

The pointer “p” always points to x as determined by compiler.

The expression “(*p)++” would increment x...

Answer 12

C.

Explanation of Solution

Modified annotated code:

long dw_loop(long x)

x in %rdi

dw_loop:

movq %rdi, %rax;Copy x to %rax

movq %rdi, %rcx

imulq %rdi, %rcx;Compute y = xx

leaq (%rdi, %rdi), %rdx;Compute n = 2x

.L2:;Loop

leaq 1(%rcx, %rax), %rax;Compute x+=y+1

subq $1, %rdx;Decrement n

testq %rdx, %rdx;Test n

jg ...

Answer 13

C.

Explanation of Solution

Modified annotated code:

long dw_loop(long x)

x in %rdi

dw_loop:

movq %rdi, %rax;Copy x to %rax

movq %rdi, %rcx

imulq %rdi, %rcx;Compute y = xx

leaq (%rdi, %rdi), %rdx;Compute n = 2x

.L2:;Loop

leaq 1(%rcx, %rax), %rax;Compute x+=y+1

subq $1, %rdx;Decrement n

testq %rdx, %rdx;Test n

jg ...

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 17

Ch. 3.4 - Prob. 3.1PP Ch. 3.4 - Prob. 3.2PP Ch. 3.4 - Prob. 3.3PP Ch. 3.4 - Prob. 3.4PP Ch. 3.4 - Prob. 3.5PP Ch. 3.5 - Prob. 3.6PP Ch. 3.5 - Prob. 3.7PP Ch. 3.5 - Prob. 3.8PP Ch. 3.5 - Prob. 3.9PP Ch. 3.5 - Prob. 3.10PP

Registers for holding values: The parameter “x” is been passed to function in register “%rdi”. Once loop is entered, there is no reference to register. The register “%rax” is initialized to “x”. The register “%rcx” is initialized to “x*x”...

Solution Summary: The author describes the registers for holding values. The loop updates the value of "x" until "n" is greater than 0.

Explanation of Solution

Registers for holding values:

The parameter “x” is been passed to function in register “%rdi”.

Once loop is entered, there is no reference to register.

The register “%rax” is initialized to “x”.

The register “%rcx” is initialized to “x*x”...

Explanation of Solution

Compiler elimination of pointer variable:

The pointer “p” always points to x as determined by compiler.

The expression “(*p)++” would increment x...

Explanation of Solution

Want to see the full answer?

Chapter 3 Solutions

Registers for holding values: The parameter “x” is been passed to function in register “%rdi”. Once loop is entered, there is no reference to register. The register “%rax” is initialized to “x”. The register “%rcx” is initialized to “x*x”...

Solution Summary: The author describes the registers for holding values. The loop updates the value of "x" until "n" is greater than 0.

Explanation of Solution

Registers for holding values: The parameter “x” is been passed to function in register “%rdi”. Once loop is entered, there is no reference to register. The register “%rax” is initialized to “x”. The register “%rcx” is initialized to “x*x”...

Explanation of Solution

Compiler elimination of pointer variable: The pointer “p” always points to x as determined by compiler. The expression “(*p)++” would increment x...

Explanation of Solution

Modified annotated code: long dw_loop(long x) x in %rdi dw_loop: movq %rdi, %rax;Copy x to %rax movq %rdi, %rcx imulq %rdi, %rcx;Compute y = x*x leaq (%rdi, %rdi), %rdx;Compute n = 2*x .L2:;Loop leaq 1(%rcx, %rax), %rax;Compute x+=y+1 subq $1, %rdx;Decrement n testq %rdx, %rdx;Test n jg ...

Want to see the full answer?

Chapter 3 Solutions

Registers for holding values:

The parameter “x” is been passed to function in register “%rdi”.

Once loop is entered, there is no reference to register.

The register “%rax” is initialized to “x”.

The register “%rcx” is initialized to “x*x”...

Compiler elimination of pointer variable:

The pointer “p” always points to x as determined by compiler.

The expression “(*p)++” would increment x...