1. CPEG323 Quiz 1 Review
Long Chen
October, 26th, 2005

2. Problem 1 (30 points) - 1 (D) MIPS is using a RISC ISA
Represent 256 in its binary representation
(C) 2^8 => [0, 255], so we need 9 bits
Big endian vs. little endian
(A) 0x (see slides)
What is Addressing mode?
(C) It determinates where an operand in an instruction can be located and addressed (see slides)

3. This number is used to identify the distance in WORDs!
Problem 1 (30 points) - 2
The range of addresses in the conditional branch instructions in MIPS
(Homework #2 - 5)
This number is used to identify the distance in WORDs!
Conditional branch instruction:
beq $s0, $s2, there 16 5
6 bits
16 bits 18
5 bits
The range of the distance is about 2^15 WORDs before and 2^15 WORDs after the branch.
In other words, 2^17 bytes before and 2^17 bytes after the branch
(D) Addresses up to about 128K before and 128K after the branch

4. Problem 1 (30 points) - 3
f) In a N-bit ALU, overflow can be detected by looking at
(A) the carry into MSB and the carry out MSB (See Topic3b)
g) -0.75ten = = -0.11two
Normalized it into -1.1twoX2-1
Compared with the general representation
(-1)s X (1+Fraction) X 2(Exponent – 127)
We have,
s= 1, Fraction = two, Exponent = 126ten = two
(A)

5. Problem 1 (30 points) - 4 Compare signed and unsigned numbers
Represent decimal number to hexadecimal number
(2) 14D
Translate IEEE 754 binary representation to decimal representation
(B) -5.0

6. Problem 2 (30 points) - 1 Homework #2 – 5 Homework #3 – 3
Caller save register: the registers that the calling procedure (caller) is responsible for saving and restoring across the call. The called procedure (callee) can then modify the registers without constraint.
Callee save register: the registers that the callee is responsible for saving and restoring if it might use. The caller uses the registers without worrying about restoring them after a call.
(see slides)

7. Problem 2 (30 points) - 2 P + P = N; N + N = P; P – N = N;
N – P = P (see slides)
i, j, temp, $fp, $ra

8. Problem 3 (10 points) Homework #3 – 1 (with slight change)
(a) unsigned INT
(b) 2’s comp INT
(c) IEEE 754 single precision FP -7.5

9. Problem 4 (15 points) - 1 Homework #3 - 6 fib(4) fib(3) fib(2) fib(2)

10. Problem 4 (15 points) - 2 sw $s0,4($sp) sw $s1,8($sp) sw $ra,0($sp)

11. Problem 5 (15 points)
Homework #3 – 2