Exam 1 Solutions
15-213 / 18-243 Fall 2010
1-b 2-a 3-c 4-b 5-a 6-d 7-b 8-a 9-a 10-a
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13-IA32: 1, 4, 4, 4 x86-64: 1, 4, 8, 8
14-a 15-a 16-a 17-a 18-a 19-elided (ambiguous) 20-a
Value FP bits Rounded value
1 011 00 1 easy normalized, exact (1 pt)
12 110 10 12 normalized exact (1 pt)
11 110 10 12 round up to 12 (because 10 is odd) (1 pt)
1/8 000 10 1/8 denorm, exact (2 pt)
7/32 001 00 1/4 straddles norm/denorm boundary, round up
Version A: H = 15 J = 3
Version B: H = 5 J = 7
abccddddeeeefXXX
There are multiple correct answers. In general, put
the largest data items first. E.g.,
ddddeeeeffffccab
A. . Inky
C. . Blinky
Cases 210, 214, and 218 should have “break”.
0x400590: 0x400470
0x400598: 0x40048a
0x4005a0: 0x40048a
0x4005a8: 0x400477
0x4005b0: 0x40047c
0x4005b8: 0x40048a
0x4005c0: 0x400482
0x4005c8: 0x40048a
0x4005d0: 0x400482
0x4005d8: 0x400487
A. (a) Call pushes the return address on the stack, jmp does not.
(b) Ret gets the return address from the top of the stack.
C. string “0123456” is stored at 0x80484d0
D. buf[0] = 0x33323130
buf[1] = 0x00363534
buf[2] = 0xffffd3e8
buf[3] = 0x080483fc
buf[4] = 0x080484d0
E. Value at %ebp is 0xffffd3e8
F. Value at %esp is 0x080483fc
G. No segfault. “0123456” is only 8 bytes including ‘\0’, and
int[2] can store 8 bytes.
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