Exam 1 Version 2 Solutions CS 213 Spring 2011 ********* Problem 1 ********* 1. c 2. b 3. d 4. b 5. c 6. b 7. d 8. c 9. b 10. a 11. a 12. T ********* Problem 2 ********* A. 67 = 0100 0011 -35 = 1101 1101 B. /* * reverseBytes - reverse bytes * Example: reverseBytes(0x12345678) = 0x78563412 * Legal ops: ! ~ & ^ | + << >> */ int reverseBytes(int x) { int newbyte0 = (x >> 24) & 0xff; int newbyte1 = (x >> 8) & 0xff00; int newbyte2 = (x << 8) & 0xff0000; int newbyte3 = x << 24; return newbyte0 | newbyte1 | newbyte2 | newbyte3; } C. x = y = 0 None x = 0 None x = 0, y = 1 ********* Problem 3 ********* A. 3 B. 15 C. 1/32 D. Bits Value Bits Value 011 000 1 101 010 5 111 000 17 111 010 NaN 110 001 9 000 011 3/32 110 010 9 1/2 110 000 8 1/2 ********* Problem 4 ********* A. Note: The correct alignment of 16-byte types on 64-bit Linux is 16 bytes. Since the cheat sheet we handed out gave incorrect information, we accepted solutions assuming 8-byte alignment also. aaaXbbbbbbXXXXXX ccccccccXXXXXXXX dddddddddddddddd eeeeeeeeffffXXXX B. dddddddddddddddd cccccccceeeeeeee ffffbbbbbbaaaXXX C. 19 D. 3 ********* Problem 5 ********* int mystery(int (*f)(int, int), int* arr, int c) { int i, x; if(c <= 0) return c; x = arr[0]; for (i = 1; i < c; i++) x = f(arr[i], x); return x; } A. The instruction places the value of %ebx onto the stack, and it decrements %esp. %ebx is a callee-saved register; since lolwut uses %ebx, its value upon invocation must be saved so it can be restored before returning to the caller. B. 0xffff0014 ********* Problem 6 ********* ----------- 0x1000 | b | <- Start argument build area here | a | | ret addr| ebp -> | %ebp | | | | | | | esp -> | | ********* Problem 7 ********* A. A[0] A[1] A[2] A[3] A[4] A[5] A[6] A[7] A[120] A[121] A[122] A[123] A[124] A[125] A[126] A[127] B. A[0] A[1] | A[120] A[121] A[2] A[3] | A[122] A[123] A[124] A[125] | A[4] A[5] A[126] A[127] | A[6] A[7]