Question

B.24 [15] <§B.5> 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 -7 > 6, which is clearly wrong. Hence, we must factor in overflow in the decision. Modify the 1-bit ALU in Figure B.5.10 on page B-33 to handle slt correctly. Make your changes on a photocopy of this figure to save time.

Short answer

The required results are tabulated below-

Overflow	The outcome	Less Than
0	0	0
0	1	1
1	0	1
1	1	0

Step-by-step solution

Define the concept.

The ALU (arithmetic logic unit) can able to give support to set less than(slt) that stores the “signed bit”. And if the value range between the ${\text{-7}}_{\text{ten}}{\text{to6}}_{\text{ten}}$then it is taken for setting less than the operation the outcome will be$-7>6$ But -7 is less than the 6. Hence, it is needed to modify the arithmetic logic unit for control the decision of overflow.

Determine the calculation.

• If there is no existence of the overflow. Then the sign bit will be used as the result. If there is an existence of the overflow, then the sign bit is wrong and inverse of sign bit will be taken.

Overflow	The outcome	LessThan
0	0	0
0	1	1
1	0	1
1	1	0

• The ALU (arithmetic logic unit) can able to give support to set less than (slt) that stores the “signed bit”.

• If the value range between the${\text{-7}}_{\text{ten}}{\text{to6}}_{\text{ten}}$ then it is taken for setting less than the operation the outcome will be role="math" localid="1655305123618" $-7>6$. But -7 is less than the 6. Hence, it is needed to modify the arithmetic logic unit for control the decision of overflow.