Segment Offset Calculation

Mapping out the offset

OK, let's break this down. Multiplying the segment by 16, subtracting it from the physical address, and working carefully with hex subtraction to refine the offset calculation.

Subtracting in hex

I’m aligning digits to subtract nibble by nibble. Borrowing as needed, the operation yields the final offset calculation: B - 0 = B, 7 - F = E.

Confirming calculation steps

I’m re-evaluating the subtraction of A37B and 0FF0, ensuring each digit aligns and any required borrowing is properly accounted for to ascertain the accurate offset.

Recalculating and aligning

I’m working through recalculating the subtraction of 0xA37B and 0xFF0, aligning digits and borrowing where necessary. The final offset, as confirmed by all methods, is 0x938B.

Clarifying the process

I’m aligning nibbles to ensure accurate subtraction of 0xA37B and 0xFF0. Confirming the final offset as 0x938B through detailed calculations and consistent methods.

Verifying accuracy

I’m breaking down the subtraction of 4A37B and 123B, ensuring each digit is correct and resulting in 49140h. This confirms the segment after addressing hex details.

Thought about memory address computation for 13 seconds

In the x86 real-mode segmented addressing scheme, the physical address is computed as:

Using this formula:

Part I: Find the Offset Given Physical = 4A37Bh and Segment = 40FFh

We rearrange the formula to solve for the offset:

1. First, compute $\text{Segment} \times 16$:

   $$40FF_{h} \times 10_{h} = 40FF0_{h}$$

2. Subtract this product from the physical address:

   $$\text{Offset} = 4A37B_{h} - 40FF0_{h}$$
   • Compute the subtraction step-by-step: $$4A37B_{h} - 40FF0_{h} = (4A37B_{h} - 40FF0_{h})$$ Subtracting: $$= A37B_{h} - 0FF0_{h}$$ Now subtract $0FF0_{h}$ from $A37B_{h}$: $$\begin{array}{r} A\;3\;7\;B \\ -\;0\;F\;F\;0 \\ \hline 9\;3\;8\;B \end{array}$$
     • $B_{h} - 0_{h} = B_{h}$
     • $7_{h} - F_{h}$: borrow 1 → $7 + 16 = 23_{h}$, $23_{h} - F_{h} = 8_{h}$ (with borrow)
     • After borrowing, $3_{h}$ becomes $2_{h}$. Now $2_{h} - F_{h}$: borrow again → $2 + 16 = 18_{h}$, $18_{h} - F_{h} = 3_{h}$
     • Borrowing makes the next digit $A_{h}$ become $9_{h}$. Then $9_{h} - 0_{h} = 9_{h}$.

   Thus:

   $$\text{Offset} = 938B_{h}$$

Part II: Find the Segment Given Physical = 4A37Bh and Offset = 123Bh

We rearrange the formula to solve for the segment:

1. Subtract the offset from the physical address:

   $$4A37B_{h} - 123B_{h}$$

   Perform the subtraction:

   $$\begin{array}{r} 4\;A\;3\;7\;B \\ -\;\;\;1\;2\;3\;B \\ \hline 4\;9\;1\;4\;0 \end{array}$$
   • $B_{h} - B_{h} = 0_{h}$
   • $7_{h} - 3_{h} = 4_{h}$
   • $3_{h} - 2_{h} = 1_{h}$
   • $A_{h} - 1_{h} = 9_{h}$
   • The most significant digit remains $4_{h}$

   So:

   $$4A37B_{h} - 123B_{h} = 49140_{h}$$

2. Now divide by 16:

   $$\text{Segment} = \frac{49140_{h}}{10_{h}} = 4914_{h}$$

Answers:

• I. The offset is 938B_1_6.
• II. The segment number is 4914_1_6.