1.Which situation will always cause the MIPS CPU's ALU overflow flag to be set to 1?

Subtracting 0 - 0

Adding 1 + 1

All of the bits of both of the ALU's inputs are inverted before the bits are added by the ALU's adder.

2.If RISC stands for Reduced Instruction Set Computing, which one of the following is an example of a RISC instruction?

# this is a MIPS assembly language comment

anArray: .word 0, 5, -2, 3

.data

add $t0, $t1, $t2

None of the above answers are correct

Adding 2147483647 + 2147483647

3.How many MIPS unsigned 32 bit integers can fit inside a MIPS CPU's general purpose register?

4

3

1

2

4.Which of the following can be no more than a byte in length for MIPS?

signed integer value

input values to the ALU’s logic operations

MIPS RAM memory address

floating point value

None of the other answers are correct

5.Overflow can occur in which situation?

a group of bits of a smaller size is represented in another group of bits in a larger size (e.g., a byte is represented inside a word)

binary numbers are subtracted from each other with an answer that is less than the number of bits representing the binary numbers

a bit is changed to its complement

a value is represented by more bits than are needed to accurately represent the value

a value needs to be represented but cannot because there are not enough bits available for the representation

6.Which of the following statements are true about aligning memory with MIPS RAM and general purpose registers?

The size of a register is equal to the size of a RAM memory location.

It takes a bit length of 4 times the number of bits a register can hold to fill the entire contents of a single RAM memory location.

7.Which of the following is not 32 bits long in the MIPS architecture?

PC

All of the other answers are correct

$t0 register

ASCII Character

RAM memory address

A register can be accessed by its 32 bit address. A RAM memory location is accessed by only 8 out of 32 bits, where 24 of the 32 bits are sign extended with the value 0.

The contents of a register cannot fit in only one RAM memory location. It takes 4 RAM memory locations to store the contents of a register.

The contents of a register are byte addressable. The contents of a RAM memory locations are not byte addressable.

8.A negative binary numeric value is guaranteed to have which one of the following

little endian byte ordering

the most significant bit value equal to 1

the least significant bit value equal to 0

All of the other answers are correct

Answer rating: 100% (QA)

1 Adding 1 1 2 add t0 t1 t2 3 1 4 MIPS RAM memory address 5 a value is represented by more bits than are needed to accurately represent the value 6 Th... View the full answer