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

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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 ...

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Check out a sample textbook solution

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Whentheuserenters!!,themostrecentcommandinthehistoryisexecuted.In the example above, if the user entered the command: Osh> !! The ‘ls -l’ command should be executed and echoed on user’s screen. The command should also be placed in the history buffer as the next command. Whentheuserentersasingle!followedbyanintegerN,theNthcommandin the history is executed. In the example above, if the user entered the command: Osh> ! 3 The ‘ps’ command should be executed and echoed on the user’s screen. The command should also be placed in the history buffer as the next command. Error handling: The program should also manage basic error handling. For example, if there are no commands in the history, entering !! should result in a message “No commands in history.” Also, if there is no command corresponding to the number entered with the single !, the program should output "No such command in history."

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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

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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

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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

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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

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Answer 16

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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...