Computer Organization and Structure
1. Convert 512ten, -1,023ten, and -4,000,000ten into 32-bit two’s complement binary numbers, respectively, and convert the following two’s complement binary numbers to be decimal numbers:
a. 1111 1111 1111 1111 1111 1110 0000 1100two;
b. 1111 1111 1111 1111 1111 1111 1111 1111two;
c. 0111 1111 1111 1111 1111 1111 1111 1111two.
2. Find the shortest sequence of MIPS instructions to determine the absolute value of a two’s complement integer. Convert this instruction (accepted by the MIPS assembler):
abs $t2, $t3
This instruction means that register $t2 has a copy of register $t3 if register $t3 is positive, and the two’s complement of register $t3 if $t3 is negative. (Hint: It can be done with three instructions.)
3. The ALU supported set on less than (slt) using just the sign bit of the adder. Let’s try a set on less than operation using the values -7ten and 6ten. To make it simpler to follow the example, let’s limit the binary representations to 4 bits: 1001two and 0110two.
1001two – 0110two = 1001two + 1010two = 0011two
This result would suggest that -7ten > 6ten, which is clearly wrong. Hence we must factor in overflow in the decision. Modify the 1-bit ALU in the following figures to handle slt correctly.
Figure 1: A 1-bit ALU that performs AND, OR, and addition on a and b or b’.
Figure 2: A 1-bit ALU for the most significant bit.
4. Add 2.85ten x 103 to 9.84ten x 104 and add 3.63ten x 104 to 6.87ten x 103, respectively, assuming that you have only three significant digits, first with guard and round digits and then without them.
5. Show the IEEE 754 binary representation for the floating-point number 10ten, 10.5ten, 0.1ten, and -2/3, respectively.