Question

Section B - ARM Assembly Language (25 marks) An ARM instruction set summary is provided at the end of this paper 1. (5 marks) Consider the following assembly instruction STMFD r13!, (r5-6} Before executing this instruction, registers hold the following values: Register Value Register r9 Value r4 0x00400040 0x00000000 r5 r10 0x11223344 0x00800080 r6 0x55667788 r11 0x10001000 r7 0x99aabbcc r12 0x20002000 r8 exddeeff00 r13 ex40004000 What memory locations are affected after executing the above instruction? In a table, with a "memory location" column and a "stored data" column, show the byte by byte values stored at each affected memory location (pay careful attention to data endianness) Additionally, what is the effect of this instruction on registers? Only write the name and new value of the registers that change 2. (10 marks) The following is the assembly language translation of a C function. mystery r1, #e cmp bne .L8 re, #e 1r myn bx .L8: r2, #e r12, [re] r3, г2 mov ldr mov .L3: ldr r2, [r0, r2, asl #2] r3, r3, #1 r12, r2 r12, r2 r3, r1 r2, r3 add cmp movcc cmp mov bne .L3 re, r12 Ir mov bx Annotate each line of the function with comments to explain what each line does. Write an equivalent C function and explain what it does. No gotos are allowed in C code. 3. (5 marks) Write a single instruction that would subtract 5 from r1 (i.e., in C r1--5), only if the result of last comparison instruction was an "greater than or equal". This instruction must not change the flags. unsigned C 4. (5 marks) Suppose that r1-0xffffffff and r2-0x2. After an "add rl,r2" instruction in Thumb-2 mode, what will be the values of r2, the negative (N) flag and the overflow (V) flag?

Answer 1

Answer :- 1) STMFD will store multiple register value to the memory address present in first operand. Here values in register r5 and r6 will be stored from memory address 0x40003FFF . The memory and value table has been shown below.

memory location	stored data
0x40003FFF	0x88
0x40003FFE	0x77
0x40003FFD	0x66
0x40003FFC	0x55
0x40003FFB	0x44
0x40003FFA	0x33
0x40003FF9	0x22
0x40003FF8	0x11

After execution, r13 = 0x40003FF7. Registers r5 and r6 are unchanged.

Answer :- 3) The instruction is-

SUBGE r1,r1,5 ; ri = r1-5, if last comparision instruction was unsigned