Question

Problem 2 Consider the following source code, where R, S, and T are constants declared with #define. long int AIRISJIT int store_ele(int h, int i, int j, long int 'dest) Alh][i]01-.dest; return sizeof(A) In compiling this program, GCC generates the following assembly code (with-02) store_ ele: movslq %esi, %rsi movslq %edi, %rdi movq (%rcx), %rax leaq (%rdi,%rdi,4), %rdi leaq (%rsi,%rsi,4), %rcx movslq %edx, %rdx addq %rcx, %rdx movq %rax, A( %rdx,8) movl $1120, %eax ret Create a text file to show your work and the steps required to determine, R, S, and T Use your reverse engineering skills to determine the values of R, S, and T, based upon the assembly code Be sure to show all the calculations required to determine R, S, and T. (15 points each for R. S. T, 15 points for steps/calculations)

Answer 1

For an array declared as Type DIRI[C], array element DO) is at address where L is the size of the data type in bytes. A. It follows that for an array declared as Type DIRI[SIT array element D)[k] is at address B. i is at 6ebp+8, j is at +12, k is at +16, dest is at+20

movl 12(%ebp), %edx #moves j to %edx leal (%edx, %edx, 4), %eax #multiplies %edx times 5, stores in %eax leal (%edx, %eax, 2), %eax #multiplies %eax times 2 and adds %edx (stores in %eax) imull S99, 8(%ebp), %edx #multiplies 99tx and stores it in %edx addl %edx, %eax #adds %edx to %eax addl 16(%ebp), %eax #adds k to %eax movl A(,%eax, 4), %edx #multiplies %eax" 4 and adds A movl 20(%ebp), %eax #moves dest into %eax movl %edx, (%eax) #moves %edx to the address pointed by %eax (dest) movi S1980, %eax #moves 1980 into %eax eax dest edx = A + 4(11j + 99x + k) Registers look like this before the last 2 lines are executed. xd + L(R(S"i + j) + k) = A + 4(11j + 99x + k)

xd = A L=4 (R(9% + j) + k) R=11 s=9 (99x + 11j + k) = (11(9x + j) + k) Since it returns 1980 as sizeof, we know the size of the array is 1980 bytes which is L(R S T) 1980 = 4(11*9*T) = 396T T=5 RESULTS R=11 T 5