Solution 00000000 e92d40f8 push r3 r4 r5 r6 r7 lr 4 e1a04000 mov r4 r0 8 e1a05001 mov r5 r1 c e1a00080 lsl r0 r0 1 10 e2800001 add r0 r0
Solution e d f push r r r r r lr e a mov r r e a mov r r c e a
Solution e d f push r r r r r lr e a mov r r e a
r r r r lr e a mov r r e a mov r r c e a lsl r r e add r r
Solution e d f push r r r r r lr e a mov r
r e a mov r r c e a lsl r r e add r r
Solution e d f push r r r r r lr e
Solution e d f push r
(Solution) 00000000 : e92d40f8 push {r3, r4, r5, r6, r7, lr} 4: e1a04000 mov r4, r0 8: e1a05001 mov r5, r1 c: e1a00080 lsl r0, r0, #1 10: e2800001 add r0, r0,...

Category: General
Words: 1050
Amount: $12
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00000000 :0: e92d40f8 push {r3, r4, r5, r6, r7, lr}4: e1a04000 mov r4, r08: e1a05001 mov r5, r1c: e1a00080 lsl r0, r0, #110: e2800001 add r0, r0, #114: e2846001 add r6, r4, #118: e1a06086 lsl r6, r6, #11c: e7913100 ldr r3, [r1, r0, lsl #2]20: e3530000 cmp r3, #024: 03a07000 moveq r7, #028: 0a000001 beq 342c: ebfffffe bl 030: e1a07000 mov r7, r034: e7953106 ldr r3, [r5, r6, lsl #2]38: e3530000 cmp r3, #03c: 03a00000 moveq r0, #040: 0a000002 beq 5044: e1a00006 mov r0, r648: e1a01005 mov r1, r54c: ebfffffe bl 050: e3540000 cmp r4, #054: 05953000 ldreq r3, [r5]58: 00877003 addeq r7, r7, r35c: 00870000 addeq r0, r7, r060: 10870000 addne r0, r7, r064: e8bd80f8 pop {r3, r4, r5, r6, r7, pc}(a) Fill in the blanks in the following C decompilation of this code based on the assembly:int canefunc(int n, int *a) {int b = ____________;int c = ____________;int d = 0;int e = 0;if (__________)d = ______________;if (__________)e = ______________;if (___________)return (_____________);return (__________);}[10] (b) What does this function do?(c) For the previous question, draw the stack frame for three recursive calls to the function:(d) Where (if at all) is the variable n stored on the stack? If it is stored, why? If it isn't stored, why?[5] (e) Explain lines 38, 3c, and 40. Describe how the N,C,V, and Z ags are set and used in theseoperations

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