ARM汇编语言面试题

1. What registers are used to store the program counter and linker register?

A:r15 and r14 are used to store the program counter and linker register, respectively.

1. What is r13 ofter used to store?

A:r13 are often used to store Stack Pointer

1. Which mode, or modes has the fewest available number of registers available? How many and why?

A: ARM has six operating modes, there are User、FIQ、IRQ、Supervisor、Abort and Undef.ARM Architecture version 4 also hasan operating mode called System. the modes of user and system can access the least registers, they can only access 17 registers, including r0-r15 and CPSR. the two modes can not access SPSR.

1. Convert the _G_C_D algorithm given in this flowchart into 1）“normal”assembler, where only branches can be conditional. 2) ARM assembler, where all instructions are conditional, thus improving code density.

“Normal” Assembler

mov r0, #27

mov r1, #9

g_c_dcmp r0, r1

beq stop

blt less ;if r0 < r1(lt表示带符号数小于)

sub r0, r0, r1

bal g_c_d ; al－always

lesssub r1, r1, r0

bal gcd

stop

ARM conditional assembler

gcd cmp r0, r1

subgt r0, r0, r1

sublt r1, r1, r0

bne gcd

1. Specify instructions which will implement the following:

a) r0 = 16 b) r1 = r0 *4

c) r0 = r1/16 d) r1 = r2 * 7

A: a) MOV r0, #16 b) MOV r1, r0, LSL #2

c) MOV r0, r1, ASR #4 d) RSBr1, r2, r2, LSL #3

1. What will the following instructions do?

a) ADDS r0, r1, r1, LSL #2 b) RSB r2, r1, #0

A: a) r0 = r1 + r1 * 4 = r1 *5 and update the conditional flags

b) r2 = 0 – r1

1. What does the following instruction sequence do?

ADD r0, r1, r1, LSL #1

SUBr0, r0, r1, LSL #4

ADD r0, r0, r1, LSL #7

A:r0 = r1 * 115 = r1 * (128 – 13) = r1 * (128 – 16 + 3) = r1 * 3- r1 * 16 + r1 * 128

8．rite a segment of code that add together elements x to x+(n-1) of an

array, where the element x = 0 is the first element of the array. Each element of the array is word size(ie. 32bits). The segment should use post-indexed addressing.At the start of your segment, you should assume that:

r0 points to the start of the array, r1 = x, r2 = n

A: ADD r0, r0, r1, LSL #2 ;set r0 to the address of element x

ADD r2, r0, r2, LSL #2 ;set r2 to the address of element x + n

MOV r1, #0 ;initialize the counter

loop

LDR r3, [r0], #4 ;access the element and mov to the next

ADD r1, r1, r3 ;add content to the counter

CMP r0, r2 ;reach element x+n?

BLT loop ;If not –repeat for next element

;on exit, sum contained in r1

9．The contents of registers r0 to r6 need to be swapped around thus:

r0 moved into r3

r1 moved into r4

r2 moved into r6

r3 moved into r5

r4 moved into r0

r5 moved into r1

r6 moved into r2

Write a segment of code that use full descending stack operations to carry this out, and hence requires no use of any other registers for temporary storage.

A: STMFD sp!,{r0-r6}

LTMFD sp!, {r3, r4, r6}

LTMFD sp!, {r5}

LTMFD sp!, {r0-r2}

10. Write a short code segment that performs a mode change by modifying the

contents of the CPSR

The mode your should change to is use mode which has the value 0x10

This assume that the current mode is a privileged mode such as supervisor mode

This would happen for instance when the processor is reset – reset code woulud be run in supervisor mode which would then need to switch to usr mode before calling the main routine in your application

You will need to usr MSR and MRS, plus 2 logical operations

A:

mmask EQU0x1f

usermEQU 0x10

#Start of here in supervisor mode

MRSr0, cpsr

BIC r0, r0, #mmask

ORRr0, r0, #userm

MSR cpsr, r0

#End up here in user mode