Given assembly code: x in %rdi fun_b: movl $64, %edx movl $0, %eax .L10: movq %rdi, %rcx andl $1, %ecx addq %rax, %rax orq %rcx, %rax shrq %rdi subq $1, %rdx jne .L10 rep; ret Data movement instructions: The different instructions are been grouped as “instruction classes”. The instructions in a class performs same operation but with different sizes of operand. The “Mov” class denotes data movement instructions that copy data from a source location to a destination. The class has 4 instructions that includes: movb: It copies data from a source location to a destination. It denotes an instruction that operates on 1 byte data size. movw: It copies data from a source location to a destination. It denotes an instruction that operates on 2 bytes data size. movl: It copies data from a source location to a destination. It denotes an instruction that operates on 4 bytes data size. movq: It copies data from a source location to a destination. It denotes an instruction that operates on 8 bytes data size. Unary and Binary Operations: The details of unary operations includes: The single operand functions as both source as well as destination. It can either be a memory location or a register. The instruction “incq” causes 8 byte element on stack top to be incremented. The instruction “decq” causes 8 byte element on stack top to be decremented. The details of binary operations includes: The first operand denotes the source. The second operand works as both source as well as destination. The first operand can either be an immediate value, memory location or register. The second operand can either be a register or a memory location. Jump Instruction: The “jump” instruction causes execution to switch to an entirely new position in program. The “label” indicates jump destinations in assembly code. The “je” instruction denotes “jump if equal” or “jump if zero”. The comparison operation is performed...

Question

Want to see the full answer?

Answer 1

Expert Solution & Answer

Chapter 3.6, Problem 3.28PP

a.

Explanation of Solution

Given assembly code:

x in %rdi

fun_b:

movl $64, %edx

movl $0, %eax

.L10:

movq %rdi, %rcx

andl $1, %ecx

addq %rax, %rax

orq %rcx, %rax

shrq %rdi

subq $1, %rdx

jne .L10

rep; ret

Data movement instructions:

The different instructions are been grouped as “instruction classes”.

The instructions in a class performs same operation but with different sizes of operand.

The “Mov” class denotes data movement instructions that copy data from a source location to a destination.

The class has 4 instructions that includes:

movb:

It copies data from a source location to a destination.

It denotes an instruction that operates on 1 byte data size.

movw:

It copies data from a source location to a destination.

It denotes an instruction that operates on 2 bytes data size.

movl:

It copies data from a source location to a destination.

It denotes an instruction that operates on 4 bytes data size.

movq:

It copies data from a source location to a destination.

It denotes an instruction that operates on 8 bytes data size.

Unary and Binary Operations:

The details of unary operations includes:

The single operand functions as both source as well as destination.

It can either be a memory location or a register.

The instruction “incq” causes 8 byte element on stack top to be incremented.

The instruction “decq” causes 8 byte element on stack top to be decremented.

The details of binary operations includes:

The first operand denotes the source.

The second operand works as both source as well as destination.

The first operand can either be an immediate value, memory location or register.

The second operand can either be a register or a memory location.

Jump Instruction:

The “jump” instruction causes execution to switch to an entirely new position in program.

The “label” indicates jump destinations in assembly code.

The “je” instruction denotes “jump if equal” or “jump if zero”.

The comparison operation is performed...

b.

Explanation of Solution

Initial test requirements:

The transformation guarded-do generate the code.

The compiler checks value of “i...

c.

Explanation of Solution

Explanation of function:

The bits in “x” are reversed using the code.

It creates a mirror image pattern of original bits...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Question 2 Using the incomplete programming code given, complete the code using dynamic programming with memory function, to reproduce the results in the following Table 1. (C++) #include<iostream>using namespace std; // max knapsack capacity // *** WRITE YOUR CODE HERE ***// num of items // *** WRITE YOUR CODE HERE ***// weight of each item // *** WRITE YOUR CODE HERE ***// value of each item // *** WRITE YOUR CODE HERE ***// variable for dynamic programming matrix // *** WRITE YOUR CODE HERE *** //==========================================// Dynamic programming function: recursive// ========================================= // ALGORITHM F(i,j) // int value // if F[i,j] is not filled yet (-1): // (start with j = W, i = n) // if capacity j < current item's weight w[i]: // value = recall F(i-1, j) // else: // we can include current item,…

(a) Assume that five generation units with third order cost function (F, (R) = A; P+ B;P+C; P; + D;) are in the circuit. Write a computer program using any abitrary programming longuage (MATLAB, C++, C#, Python,.) to calculate economic load dipatch (ELD) using first order gradient method. Note that all parameters and variables should be defined inside the program (at tirst lines) such that units' characteristics and demand can be changed easily. Neglect grid losses. jusing dynamic programming (DP) methed.

Problem (7) « :(F'ind 3 possible places for errors in the following code (assume no issues with parts not shown, such as #include). Assume no user-defined global variables exist. Explain specifically what causes the error and whether it is a syntax, runtime or logic error. int sumArray(int a[], int size); int main() { int a={1,5,6,3,7,4}; int sum = sumArray(a, 5); cout « "Sum is: " « sum <« endl; } int sumArray(int a[], int size) { if(size == 0) { return size; } else { a[size-1] + sumArray(a, size--); } int dummy = 8; }

Answer 2

Expert Solution & Answer

Chapter 3.6, Problem 3.28PP

a.

Explanation of Solution

Given assembly code:

x in %rdi

fun_b:

movl $64, %edx

movl $0, %eax

.L10:

movq %rdi, %rcx

andl $1, %ecx

addq %rax, %rax

orq %rcx, %rax

shrq %rdi

subq $1, %rdx

jne .L10

rep; ret

Data movement instructions:

The different instructions are been grouped as “instruction classes”.

The instructions in a class performs same operation but with different sizes of operand.

The “Mov” class denotes data movement instructions that copy data from a source location to a destination.

The class has 4 instructions that includes:

movb:

It copies data from a source location to a destination.

It denotes an instruction that operates on 1 byte data size.

movw:

It copies data from a source location to a destination.

It denotes an instruction that operates on 2 bytes data size.

movl:

It copies data from a source location to a destination.

It denotes an instruction that operates on 4 bytes data size.

movq:

It copies data from a source location to a destination.

It denotes an instruction that operates on 8 bytes data size.

Unary and Binary Operations:

The details of unary operations includes:

The single operand functions as both source as well as destination.

It can either be a memory location or a register.

The instruction “incq” causes 8 byte element on stack top to be incremented.

The instruction “decq” causes 8 byte element on stack top to be decremented.

The details of binary operations includes:

The first operand denotes the source.

The second operand works as both source as well as destination.

The first operand can either be an immediate value, memory location or register.

The second operand can either be a register or a memory location.

Jump Instruction:

The “jump” instruction causes execution to switch to an entirely new position in program.

The “label” indicates jump destinations in assembly code.

The “je” instruction denotes “jump if equal” or “jump if zero”.

The comparison operation is performed...

b.

Explanation of Solution

Initial test requirements:

The transformation guarded-do generate the code.

The compiler checks value of “i...

c.

Explanation of Solution

Explanation of function:

The bits in “x” are reversed using the code.

It creates a mirror image pattern of original bits...

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 3.28PP

a.

Explanation of Solution

Given assembly code:

x in %rdi

fun_b:

movl $64, %edx

movl $0, %eax

.L10:

movq %rdi, %rcx

andl $1, %ecx

addq %rax, %rax

orq %rcx, %rax

shrq %rdi

subq $1, %rdx

jne .L10

rep; ret

Data movement instructions:

The different instructions are been grouped as “instruction classes”.

The instructions in a class performs same operation but with different sizes of operand.

The “Mov” class denotes data movement instructions that copy data from a source location to a destination.

The class has 4 instructions that includes:

movb:

It copies data from a source location to a destination.

It denotes an instruction that operates on 1 byte data size.

movw:

It copies data from a source location to a destination.

It denotes an instruction that operates on 2 bytes data size.

movl:

It copies data from a source location to a destination.

It denotes an instruction that operates on 4 bytes data size.

movq:

It copies data from a source location to a destination.

It denotes an instruction that operates on 8 bytes data size.

Unary and Binary Operations:

The details of unary operations includes:

The single operand functions as both source as well as destination.

It can either be a memory location or a register.

The instruction “incq” causes 8 byte element on stack top to be incremented.

The instruction “decq” causes 8 byte element on stack top to be decremented.

The details of binary operations includes:

The first operand denotes the source.

The second operand works as both source as well as destination.

The first operand can either be an immediate value, memory location or register.

The second operand can either be a register or a memory location.

Jump Instruction:

The “jump” instruction causes execution to switch to an entirely new position in program.

The “label” indicates jump destinations in assembly code.

The “je” instruction denotes “jump if equal” or “jump if zero”.

The comparison operation is performed...

b.

Explanation of Solution

Initial test requirements:

The transformation guarded-do generate the code.

The compiler checks value of “i...

c.

Explanation of Solution

Explanation of function:

The bits in “x” are reversed using the code.

It creates a mirror image pattern of original bits...

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 3.28PP

a.

Explanation of Solution

Given assembly code:

x in %rdi

fun_b:

movl $64, %edx

movl $0, %eax

.L10:

movq %rdi, %rcx

andl $1, %ecx

addq %rax, %rax

orq %rcx, %rax

shrq %rdi

subq $1, %rdx

jne .L10

rep; ret

Data movement instructions:

The different instructions are been grouped as “instruction classes”.

The instructions in a class performs same operation but with different sizes of operand.

The “Mov” class denotes data movement instructions that copy data from a source location to a destination.

The class has 4 instructions that includes:

movb:

It copies data from a source location to a destination.

It denotes an instruction that operates on 1 byte data size.

movw:

It copies data from a source location to a destination.

It denotes an instruction that operates on 2 bytes data size.

movl:

It copies data from a source location to a destination.

It denotes an instruction that operates on 4 bytes data size.

movq:

It copies data from a source location to a destination.

It denotes an instruction that operates on 8 bytes data size.

Unary and Binary Operations:

The details of unary operations includes:

The single operand functions as both source as well as destination.

It can either be a memory location or a register.

The instruction “incq” causes 8 byte element on stack top to be incremented.

The instruction “decq” causes 8 byte element on stack top to be decremented.

The details of binary operations includes:

The first operand denotes the source.

The second operand works as both source as well as destination.

The first operand can either be an immediate value, memory location or register.

The second operand can either be a register or a memory location.

Jump Instruction:

The “jump” instruction causes execution to switch to an entirely new position in program.

The “label” indicates jump destinations in assembly code.

The “je” instruction denotes “jump if equal” or “jump if zero”.

The comparison operation is performed...

b.

Explanation of Solution

Initial test requirements:

The transformation guarded-do generate the code.

The compiler checks value of “i...

c.

Explanation of Solution

Explanation of function:

The bits in “x” are reversed using the code.

It creates a mirror image pattern of original bits...

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 3.28PP

a.

Explanation of Solution

Given assembly code:

x in %rdi

fun_b:

movl $64, %edx

movl $0, %eax

.L10:

movq %rdi, %rcx

andl $1, %ecx

addq %rax, %rax

orq %rcx, %rax

shrq %rdi

subq $1, %rdx

jne .L10

rep; ret

Data movement instructions:

The different instructions are been grouped as “instruction classes”.

The instructions in a class performs same operation but with different sizes of operand.

The “Mov” class denotes data movement instructions that copy data from a source location to a destination.

The class has 4 instructions that includes:

movb:

It copies data from a source location to a destination.

It denotes an instruction that operates on 1 byte data size.

movw:

It copies data from a source location to a destination.

It denotes an instruction that operates on 2 bytes data size.

movl:

It copies data from a source location to a destination.

It denotes an instruction that operates on 4 bytes data size.

movq:

It copies data from a source location to a destination.

It denotes an instruction that operates on 8 bytes data size.

Unary and Binary Operations:

The details of unary operations includes:

The single operand functions as both source as well as destination.

It can either be a memory location or a register.

The instruction “incq” causes 8 byte element on stack top to be incremented.

The instruction “decq” causes 8 byte element on stack top to be decremented.

The details of binary operations includes:

The first operand denotes the source.

The second operand works as both source as well as destination.

The first operand can either be an immediate value, memory location or register.

The second operand can either be a register or a memory location.

Jump Instruction:

The “jump” instruction causes execution to switch to an entirely new position in program.

The “label” indicates jump destinations in assembly code.

The “je” instruction denotes “jump if equal” or “jump if zero”.

The comparison operation is performed...

b.

Explanation of Solution

Initial test requirements:

The transformation guarded-do generate the code.

The compiler checks value of “i...

c.

Explanation of Solution

Explanation of function:

The bits in “x” are reversed using the code.

It creates a mirror image pattern of original bits...

Answer 8

Chapter 3.6, Problem 3.28PP

a.

Explanation of Solution

Given assembly code:

x in %rdi

fun_b:

movl $64, %edx

movl $0, %eax

.L10:

movq %rdi, %rcx

andl $1, %ecx

addq %rax, %rax

orq %rcx, %rax

shrq %rdi

subq $1, %rdx

jne .L10

rep; ret

Data movement instructions:

The different instructions are been grouped as “instruction classes”.

The instructions in a class performs same operation but with different sizes of operand.

The “Mov” class denotes data movement instructions that copy data from a source location to a destination.

The class has 4 instructions that includes:

movb:

It copies data from a source location to a destination.

It denotes an instruction that operates on 1 byte data size.

movw:

It copies data from a source location to a destination.

It denotes an instruction that operates on 2 bytes data size.

movl:

It copies data from a source location to a destination.

It denotes an instruction that operates on 4 bytes data size.

movq:

It copies data from a source location to a destination.

It denotes an instruction that operates on 8 bytes data size.

Unary and Binary Operations:

The details of unary operations includes:

The single operand functions as both source as well as destination.

It can either be a memory location or a register.

The instruction “incq” causes 8 byte element on stack top to be incremented.

The instruction “decq” causes 8 byte element on stack top to be decremented.

The details of binary operations includes:

The first operand denotes the source.

The second operand works as both source as well as destination.

The first operand can either be an immediate value, memory location or register.

The second operand can either be a register or a memory location.

Jump Instruction:

The “jump” instruction causes execution to switch to an entirely new position in program.

The “label” indicates jump destinations in assembly code.

The “je” instruction denotes “jump if equal” or “jump if zero”.

The comparison operation is performed...

Answer 9

Chapter 3.6, Problem 3.28PP

a.

Explanation of Solution

Given assembly code:

x in %rdi

fun_b:

movl $64, %edx

movl $0, %eax

.L10:

movq %rdi, %rcx

andl $1, %ecx

addq %rax, %rax

orq %rcx, %rax

shrq %rdi

subq $1, %rdx

jne .L10

rep; ret

Data movement instructions:

The different instructions are been grouped as “instruction classes”.

The instructions in a class performs same operation but with different sizes of operand.

The “Mov” class denotes data movement instructions that copy data from a source location to a destination.

The class has 4 instructions that includes:

movb:

It copies data from a source location to a destination.

It denotes an instruction that operates on 1 byte data size.

movw:

It copies data from a source location to a destination.

It denotes an instruction that operates on 2 bytes data size.

movl:

It copies data from a source location to a destination.

It denotes an instruction that operates on 4 bytes data size.

movq:

It copies data from a source location to a destination.

It denotes an instruction that operates on 8 bytes data size.

Unary and Binary Operations:

The details of unary operations includes:

The single operand functions as both source as well as destination.

It can either be a memory location or a register.

The instruction “incq” causes 8 byte element on stack top to be incremented.

The instruction “decq” causes 8 byte element on stack top to be decremented.

The details of binary operations includes:

The first operand denotes the source.

The second operand works as both source as well as destination.

The first operand can either be an immediate value, memory location or register.

The second operand can either be a register or a memory location.

Jump Instruction:

The “jump” instruction causes execution to switch to an entirely new position in program.

The “label” indicates jump destinations in assembly code.

The “je” instruction denotes “jump if equal” or “jump if zero”.

The comparison operation is performed...

Answer 10

b.

Explanation of Solution

Initial test requirements:

The transformation guarded-do generate the code.

The compiler checks value of “i...

Answer 11

b.

Explanation of Solution

Initial test requirements:

The transformation guarded-do generate the code.

The compiler checks value of “i...

Answer 12

c.

Explanation of Solution

Explanation of function:

The bits in “x” are reversed using the code.

It creates a mirror image pattern of original bits...

Answer 13

c.

Explanation of Solution

Explanation of function:

The bits in “x” are reversed using the code.

It creates a mirror image pattern of original bits...

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

Solution Summary: The author explains the different instructions in a class perform the same operation but with different sizes of operands.

Explanation of Solution

Explanation of Solution

Initial test requirements:

The transformation guarded-do generate the code.

The compiler checks value of “i...

Explanation of Solution

Explanation of function:

The bits in “x” are reversed using the code.

It creates a mirror image pattern of original bits...

Want to see the full answer?

Chapter 3 Solutions

Solution Summary: The author explains the different instructions in a class perform the same operation but with different sizes of operands.

Explanation of Solution

Explanation of Solution

Initial test requirements: The transformation guarded-do generate the code. The compiler checks value of “i...

Explanation of Solution

Explanation of function: The bits in “x” are reversed using the code. It creates a mirror image pattern of original bits...

Want to see the full answer?

Chapter 3 Solutions

Initial test requirements:

The transformation guarded-do generate the code.

The compiler checks value of “i...

Explanation of function:

The bits in “x” are reversed using the code.

It creates a mirror image pattern of original bits...