Flag values: Register values: EAX = FF82E2 EBX = FFFh a) What are the flag values and new value of EAX after the following operation? ADD AH, 7Eh Z = C = OV= EAX=
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- MOV AL, 5 ADD AL, 4 The flag register is affected by the above addition and the carry flag will equal * to 1 MOV AL,OF5H ADD AL,OBH After the addition AL will contain FF BB 00 11 IK8. Show the steps involved in the following MOV AL,6EH ; AL=6EH ; BH=1CH ; subtract BH from AL. Place result in AL. MOV BH,1CH SUB AL,BH Find out the contents of the flag register.MIPS Assembly Write a program for RISC MIPS 32bits that challenges a person to guess a number of up to 3 digits before and 3 digits after the comma, that is, a value between 000,000 and 999,999. The program user will try to guess the number with multiple attempts, where the program will indicate at each attempt if the informed number is bigger, smaller, or if the user guessed correctly. When this happens, the program ends, showing a SUCCESS message and a total count of the number of attempts taken to get it right. The program must have a SUBROUTINE as described below. SUBROUTINE:- Input Parameters: user guess, expected value- Output: message on screen (bigger, smaller, SUCCESS)- Returns: value 1 if correct, 0 if incorrect
- Shift 86>>3A(n) __________ is a storage location implemented in the CPU.In X86 Assembly language: Write a procedure to check if a number is divisible by an other number Then write another procedure to set the divisor and dividend Call procedure to set values Call procedure to set eax to 1 if divisible, 0 if not divisible
- If the value of ESP register before running line number 9 is 2000. What will be its value at line 15? Note that the "sum" function is defined in the second picture below. extern sum, print_sint, print_uint, print_hex global main segment .text main: push 1 push-2 call sum add esp, 8 push eax call print_sint call print_uint call print_hex add esp, 4 Imov eax, 1 int 0x80 #include int sum(int a, int b) { return a+b; } void print_sint(int a) { printf("%d\n", a); } void print_uint (int a) { printf("%u\n", a); void print_hex(int a) { printf("%x\n", a); } A. 1996 B. 2012 C. 1988 D. 2008|long arith(long x,long y,long z) { long t1=x+y; long t2=z+t1; long t3=x+4; |long t4=y*48; long t5=t3+t4; long rval=t2*t5; return rval ; } In what registers are the arguments x, y, and z passed? In what register is variable rval returned? How many registers are involved in the execution of all the ALU operations? How many bytes are allocated to the stack frame of function arith? Justify your answer. Annotate each line of the assembly code obtained from the compiler and include the screenshot.The ADD and SUB operators affect all the status flags according to the result of the operation. Give the outputs of the following sequential operations in assembly language and determine whether overflow flag (OF) or carry flag (CF) will be signaled. al, al, al, al, al, al, mox OFFh add 1 mox 7Fh add 1 80h mox add 80h
- C++ Program: Make a program that would demonstrate the concept of Line Coding: Polar NRZ: Remember: if next bit is 1, there is inversion if next bit is 0, no inversion neg – means the signal is in the negative axis pos – means the signal is in the positive axis inv – means there is inversion no inv – means there is no inversion a. initially high (starts with positive, inverse is negative) Input: 1 1 1 0 0 1 0 1 1 0 0 0 Output: inv neg inv pos inv neg no inv neg no inv neg inv pos no inv pos inv neg inv pos no inv pos no inv pos no inv pos b. initially low (starts with negative, inverse is positive) Input: 1 1 1 0 0 1 0 1 1 0 0 0 Output: inv pos inv neg inv pos no inv pos no inv pos inv neg no inv neg inv pos inv neg no inv neg no inv neg no inv neg The logic here is when input is 1, next inputted 1 will be inverse…* Direction flag can reset to be zero by CLC True O False O ADD is a data transfer instruction which is used to add byte to byte/word to .word true O False O instruction is used to loon a set of instructions till zero flag becomesList an application where you can explain the advantage of using the Gray code over the Binary code.