x choices: ~b -b 1-b b 1+b y choices: b+res1 a+res1 a+c b+c a+b z choices: a * 1

Question

x choices:

~b

-b

1-b

b

1+b

y choices:

b+res1

a+res1

a+c

b+c

a+b

z choices:

a * 15

res1 * 23

res1 * 15

b * 17

c * 23

w choices:

res2 << 5

c << 5

a >> 5

c >> 5

b >>5

Consider the following function where X, Y, Z, and W represent incomplete code int math (int a int b, int c) int res1 X; int res Y; int res Z; int res4 W; return res4; The function above is implemented in assembly code below where the function parameters a, b and c are at the effective addresses 0x8 ebp), 0xc ebp) and 0x10 ebp), respectively. mov exc (%ebp),%eax not %eax mov %eax, 0x10(%ebp) mov 0x10(%ebp),%eax ov ex8 (%ebp),%edx add %edx,%eax mov %eax, 0xc(%ebp) mov 0x10 (%ebp),%ea imul $0x17,%eax,%eax mov %eax, -0x8 (%ebp) mov 0x10(%ebp),%eax sar $0x5,%eax mov %eax, 0x4(%ebp) Choose the correct options for X, Y, Z and W to complete the function so that it corresponds with its assembly code. X is [Select Y is Select z is [Select w is Select

Explanation / Answer

The answer is
X is ~b
Y is a+c
Z is c * 23
W is c >> 5

Explanation

a. In Line 1, we are storing value of b into eax register. Then in Line-2 we are performing not(~) operation on it. Therefore X is ~b
b. In line 4&5, we are moving values of a and c, into eax and edx registers respectively. Then on the next line, we are adding them. Therefore Y = a+c
c. In line 8 we are storing value of c into eax register. Then on line 9, we are multiplying 0x17(= 23 in decimal) with eax. Therefore, Z = c*23
d. In line 11 we are storing value of c into eax register. Then on line 12, we are performing shift right operation, by offset 5. Therefore, W = c>>5

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