Question

1. The following instruction(s),--, is (are) ARM® Cortex®-M4 assembly instructions. a. ui32int uiLoop c. void_WFI(void); d. WFI 2. The ARM® Cortex-M4 assembly instructions contain. a. Label field b. Operation field and operands field c. Operation field, op erands field and comment field d. Both a and c Copyrighted material Homework 245 3. The pseudo instruction DCW 2 is to a. Reserve 2-byte space for a data item in the program b. Reserve 2 hal f-word space for a data item in the program c. Reserve 4-byte space for a data item in the program space for a data item in the program d. Reserve 2-word 4. The LDRB RO , [R2 , #0x5 ] is to load an unsigned byte located at-_ into the register R0 if the R2 contains 0x40004000. b. 0x40004000 c. 0x40004005 d. 0x40000000 5. The instruction BFC R4, #8, #12; is to -' a. Set bits 8-12 in the register R4 to 1 b. Reset bits 8-12 in the register R4 to 0 c. Set bits 8-19 in the register R4 to 1 d. Reset bits 8-19 in the register R4 to 0

Answer 1

1))option a -There is no instruction in ARM of form of option a

FOR option b cannot be correct as ARM instructions cannot be of that form.

FOR option c and d

WFI (wait for interrupt) IS FOR ENTERING EITHER STANDBY OR SHUTDOWN MODE WHERE AN INTERRUPT IS REQUIRED TO WAKEUP THE PROCESSOR.

Even void _WFI(void) is used to enter low power state.

So C and D are correct.

So answers are C,D.

2))ARM instruction set has is load/store architecture.

It has LABEL,OPCODE,OPERANDS,COMMENTS.

LABEL-optional and used to define starting location of block of statements.

OPCODE-is the seconf field and used to tell which operation to peform on operands.

OPERANDS-number of operands varies from 0 to 4 in ARM Architecture.

COMMENTS-messages for human consumption.

so option D is correct.

3)) DCW(DEFINE CONSTANT WORD) is a data reservation directive which allocates one or more half words of memory,aligned on two-byte boundaries.

here DCW 2 is used to reserve 2-half word space.

Therefore option B is right.

4))LDRB-(load register byte)

[r2,#0x5] tells to add value in r2 and 0x5 and value at that address is loaded into r0.

Adding r2=0x40004000 and 0x5 gives 0x40004005.

So option C is correct.

5))BFC(Bit Field Clear) is used to reset bits to zero.

Also BFC R4,#y,#x where y is start,x is number of bits to reset.

Here start from 8 and then 8+12-1=19 .

So option D is right.