ARM汇编代写代考

CS1021 Tutorial 6 Reading and Writing to Memory Q1 if a, b and c are 32bit signed integers stored at memory addresses 0x40000000, 0x40000004 and 0x40000008 respectively, write ARM assembly language instructions to compute: (i) a = a + b (ii) c=a-b (iii) b=a*a Q2 If a, b and c are 256 bit integers stored

CS1021 Tutorial 2 Q1 The NXP LPC2468 has 512KiB of flash memory starting at address 0x00000000. What is the last address of the memory area (in hexadecimal)? Q2 The NXP LPC2468 has 64KiB of read write memory starting at address 0x40000000. What is the last address of the memory area (in hexadecimal)? Q3 One hundred

CS1021 Lab 1 Evaluating Expressions Assuming x is in R1, y in R2, z in R3 and the result in R0, write and test an ARM assembly language program to evaluate the following expressions: (a) 3×2 + 5x (b) 2×2 + 6xy + 3y2 (c) x3 − 4×2 + 3x + 8 (d) 3×4 –

CS1021 Tutorial 3 Q1 Translate each of the following pseudo-code statements into a sequence of ARM assembly language instructions. Assume that x and y are signed integers and that x is in R1 and y in R2. (i) if (x== 0) x = x + 5; (ii) if (x >= 5) x = 0; (iii)

CS1021 Tutorial 3 Q1 Translate each of the following pseudo-code statements into a sequence of ARM assembly language instructions. Assume x and y are signed integers and x is in R1 and y is in R2. (i) if (x== 0) x = x + 5; CMP R1,#0 BNE L1 ADD R1, R1, #5 L1 …

CS1021 Lab 2 Fibonacci Numbers Q1 Write an ARM Assembly Language program to compute the nth Fibonacci number. The nth Fibonacci number is defined recursively as: Fn =Fn-1 +Fn-2 whereF0 =0andF1 =1. Use R2 for n and compute Fn in R0. Test your program by computing F16 and F32. Q2 Using your answer to Q1

CS1021 Lab 3 Greatest Common Divisor Q1 The greatest common divisor (GCD) of two positive integers a and b is the largest integer that divides both a and b without a remainder. For example, the GCD of 24 and 32 is 8. Convert the following pseudo code for calculating the GCD of a and b

L1 MOV R2, #0 L2 ;x==0x30 ; if TRUE execute s0 ;x==0x31 ; if TRUE execute s0 ;x>=0x40 ; if FALSE execute s1 else s0 ; y = 1 (s0) ; skip s1 ; y = 0 (s1) if (((x >= 0x30) && (x = 0x41) && (x =0x30 ; if FALSE check ((x >=

ARM ASSEMBLY LANGUAGE Simple Model of an ARM Microprocessor System • comprises a central processing unit (CPU) and memory • instructions and data are stored in memory • the CPU reads instructions from memory (one after another) and executes them • when the CPU executes an instruction  it performs operations between its registers OR