Consider the following fragment of Brookshear machine assembly language code: MOV FC -> R2 // Load R2 with 11111100 (binary) AND R1, R2 -> R1 // Put R1 AND R2 in R1 ROT R1, 2 // Rotate R1 by 2 bits right If the initial value held in R1 (register 1) is interpreted as a non-negative integer, what is the programmer's overall aim in writing this fragment of code?

Consider the following fragment of Brookshear machine assembly language code: MOV FC -> R2 // Load R2 with 11111100 (binary) AND R1, R2 -> R1 // Put R1 AND R2 in R1 ROT R1, 2 // Rotate R1 by 2 bits right If the initial value held in R1 (register 1) is interpreted as a non-negative integer, what is the programmer's overall aim in writing this fragment of code?

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

See similar textbooks

Similar questions

Review the Wireshark TCP data flow shown here: Host A -- SYN --> Host B Seg =0 Ack = 919412342 Host A Host B Seq = 1 Ack = 1 Host A PSH, ACK Len: 512 --> Host B Seg = 1 Ack = 1 --- Host A Host B Seg = 901 Ack = 1342 --- Host A Host B Seq = 901 Ack = 2992 --- Host A <--- ACK Len: 1452 --- Host B Seq = 2992 Ack = 901 Which of the following statements is correct and represents the next appropriate acknowledgment from Host A? C. Sequence number 4444, Acknowledgment number 901 Selected Answer: Answers: A. Sequence number 2993, Acknowledgment number 901 B. Sequence number 901, Acknowledgment number 2993 C. Sequence number 4444, Acknowledgment number 901 Activate D. Sequence number 901, Acknowledgment number 4444 Go to Setti
Write an assembly code to assign the register BX with 4014H, and register DX with 4028 H. Move the content of BX to DX, und verse versa. Then, copy the now value of low content for BX and DX .to memory location [2131 H], and [2139H]
(a) In MIPS machine language, the R format instructions need an additional field (the “funct” field) to specify the instruction type (“add”, “sub”, etc.). Why can’t this be done using just the 6-bit “op” field, which all MIPS instructions have, instead? (b) Give a sequence of MIPS assembly language instructions that adds the contents of $s0 and $s1, storing the result into $s0, if the contents of $s1 are greater than 6. Otherwise, the contents of $s0 should be incremented by 28. In either case, afterwards the contents of $s0 should be copied into $s2. (c) Suppose that the starting address of an array of words is stored in $s0. Give a MIPS assembly language instruction that loads the contents of the second word in the array into $s1. (d) Suppose that A and B are defined using “A: .word 0”, and “B: .space 100”. Give a sequence of MIPS assembly language instructions that calls a procedure “P” with its two arguments being the contents of A and the address of B, using the standard procedure…
Explain in detail and no rush. Please ensure quality is a priority.
VDHL simple computer question.
Implement the following expression in assembly language: result = (val3 + val4) - (val1 – val2) – (30*4)/55- Assume that result,val1, val2, val3 and val4 are 16-bit integer variables- You need to implement the expression the way it is provided, you cannot do any reduction on theexpression while implementing it.- Initialize val1 with 134 (hexadecimal), val2 with 139 (hexadecimal), val3 with 67(hexadecimal) and val4 with 47 (hexadecimal)- You are NOT allowed to update the values of any variables.- Use ONLY mov, add, sub, movzx, movsx, or neg instructions whenever needed.- Use the debugger to verify your answer
E In the following code block(Reference:Q11), you will a set of assembly instructions with corresponding line numbers (line numbers are for informational purpose only and they are not part of the source code). 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 mov edx, 5 dec ecx jmp LABEL1 mov eax, 1 LABEL1: mul edx jmp ecx mov edx, 0678h sub edx, eax jmp DWORD PTR [edx] neg ebx add ecx, ebx mov eax, 0 For each of the conditions/scenario listed below, indicate the corresponding line number (that cause or is associated with the condition/scenario). Enter 0 (Zero) if the condition is not caused by the block of code. 1) Memory indirect jump: type your answer... type your answer... type your answer... type your answer... 2) Register indirect jump: 3) Relative short jump: type your answer... 5) Two's complement type your answer... 4) Relative near jump: 6) Unreachable code