Add two positive 8-bit binary integers
Data Representation – Adding Two Positive 8‑bit Binary Integers
Learning Objectives (AO1 & AO2)
- Convert accurately between decimal, binary and hexadecimal (8‑bit) representations.
- Perform binary addition of two positive 8‑bit numbers and decide whether the result fits in a byte.
- Identify and explain unsigned overflow and signed (two’s‑complement) overflow.
- Apply logical left‑ and right‑shifts as a quick way to multiply or divide by powers of two.
- Add a positive and a negative two’s‑complement number and interpret the result.
Key Concepts Overview
- Binary numeral system – base 2, digits 0 and 1.
- 8‑bit unsigned range – 0 to \(2^{8}-1 = 255\).
- 8‑bit signed (two’s‑complement) range – –128 to +127.
- Overflow terminology (as used in the Cambridge syllabus):
- Unsigned overflow: a carry out of the most‑significant bit (MSB).
- Signed overflow: the carry into the MSB differs from the carry out of the MSB (equivalently, adding two numbers with the same sign produces a result with a different sign).
- Binary addition rules (including the case of three bits being added):
| Bits added | Sum bit | Carry‑out |
|---|---|---|
| 0 + 0 | 0 | 0 |
| 0 + 1 | 1 | 0 |
| 1 + 0 | 1 | 0 |
| 1 + 1 | 0 | 1 |
| 1 + 1 + 1 | 1 | 1 |
Number‑System Conversions
Decimal → 8‑bit Binary
- Divide the decimal number by 2, record the remainder.
- Repeat with the quotient until it becomes 0.
- Read the remainders in reverse order; pad with leading 0’s to obtain exactly 8 bits.
Binary → Hexadecimal (8‑bit)
- Group the binary number into two 4‑bit nibbles (from the right).
- Convert each nibble to its hexadecimal digit (0‑F).
Hexadecimal → Binary (8‑bit)
- Write each hex digit as its 4‑bit binary equivalent.
- Combine the two groups to obtain the full 8‑bit binary pattern.
| Decimal | Binary (8‑bit) | Hexadecimal |
|---|---|---|
| 0 | 0000 0000 | 00 |
| 10 | 0000 1010 | 0A |
| 45 | 0010 1101 | 2D |
| 78 | 0100 1110 | 4E |
| 255 | 1111 1111 | FF |
Two’s‑Complement (Signed) Representation
Logical Shifts
Step‑by‑Step Procedure for Adding Two Positive 8‑bit Numbers
- Write each operand as an 8‑bit binary string. Add leading zeros if necessary.
- Align the numbers vertically** with the least‑significant bit (LSB) on the right.
- Column‑wise addition** starting at the LSB. Record the sum bit and the carry to the next column.
- Proceed to the MSB. If a carry remains after the MSB, an unsigned overflow has occurred.
- Signed‑overflow check** (two’s‑complement only): compare the carry **into** the MSB with the carry **out of** the MSB. If they differ, a signed overflow has occurred.
- Optional verification:** Convert the binary result back to decimal (or hex) to confirm.
Worked Example – Unsigned Addition (No Overflow)
Add 45₁₀ and 78₁₀.
Step 1 – Decimal → 8‑bit Binary
| Decimal | 8‑bit Binary |
|---|---|
| 45 | 0010 1101 |
| 78 | 0100 1110 |
Step 2 – Column‑wise Binary Addition
| Bit position | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|---|---|---|
| A | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| B | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 |
| Carry‑in | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| Sum | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 |
| Carry‑out | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
Result: 0111 0111₂ → \(64+32+8+4+2+1 = 111_{10}\).
Carry‑out from the MSB = 0 → no unsigned overflow. MSB = 0, so there is also no signed overflow.
Worked Example – Signed Overflow (Two’s‑Complement)
Add +120₁₀ and +100₁₀ as signed 8‑bit numbers.
| Decimal | Binary (8‑bit) |
|---|---|
| +120 | 0111 1000 |
| +100 | 0110 0100 |
Binary addition yields 1101 1100 with a carry‑out of 0.
- MSB = 1 → the result would be interpreted as negative in two’s‑complement.
- Carry into the MSB = 1, carry out = 0 → they differ → signed overflow.
- Interpreting
1101 1100₂as two’s‑complement gives –36, not the expected +220.
Worked Example – Adding a Positive and a Negative Two’s‑Complement Number
Add +45₁₀ and –20₁₀.
| Decimal | Binary (8‑bit) |
|---|---|
| +45 | 0010 1101 |
| –20 | 1110 1100 |
Column‑wise addition:
| Bit | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|---|---|---|
| A | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| B | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 |
| Carry‑in | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| Sum | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
| Carry‑out | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
Result = 1000 1001₂. Interpreting as two’s‑complement gives –31, which is indeed 45 + (‑20) = 25? Wait – correction: the correct two’s‑complement for –20 is 1110 1100, adding yields 0010 1111₂ = 47. The previous table contained a mistake; the final correct result is:
| Step | Binary |
|---|---|
| +45 | 0010 1101 |
| –20 (two’s‑complement) | 1110 1100 |
| Sum | 0001 1001 |
Result = 0001 1001₂ = 25₁₀. No overflow occurs (carry out of MSB = 0, carries into/out of MSB are equal).
Overflow Summary Table
| Type | Condition | Interpretation |
|---|---|---|
| Unsigned overflow | Carry out of the MSB = 1 | True sum > 255 (does not fit in an 8‑bit byte) |
| Signed overflow (two’s‑complement) | Carry‑into MSB ≠ Carry‑out MSB | Result cannot be represented in the range –128 … +127 |
Check‑Your‑Understanding Boxes (AO2)
Box 1 – Hexadecimal to Binary
Convert
Convert
0x3A to an 8‑bit binary number.
Box 2 – Unsigned Overflow Detection
Add
Add
1101 1010₂ (218) and 0110 1111₂ (111). State whether unsigned overflow occurs and give the 8‑bit result.
Box 3 – Signed Overflow Explanation
Explain why adding
Explain why adding
0111 1110₂ (+126) and 0000 0011₂ (+3) causes signed overflow, even though the unsigned carry‑out is 0.
Box 4 – Logical Shift Multiplication
Shift
Shift
0010 1101₂ (45) left by 2 positions. Write the binary result, the hexadecimal equivalent, and the decimal value it now represents.
Box 5 – Adding a Positive and a Negative Number
Add
Add
+45₁₀ and –20₁₀ using two’s‑complement. Show each step and state whether any overflow occurs.
Practice Questions (Full‑Mark AO1/AO2)
- Add 120₁₀ and 95₁₀** using 8‑bit binary addition. Indicate:
- Unsigned overflow? (yes/no)
- Signed overflow? (yes/no)
- Provide the 8‑bit binary result and its hexadecimal form.
- What is the maximum unsigned value that can be stored in an 8‑bit byte? Write it in decimal, binary and hexadecimal.
- Convert –45₁₀ to an 8‑bit two’s‑complement binary number and then back to decimal to verify.
- Perform a logical left shift of
0010 1101₂by two positions. What decimal value does the result represent? State whether overflow occurs. - Explain, with reference to the examples above, why a carry‑out from the MSB guarantees unsigned overflow but does **not** always indicate signed overflow.
Suggested Diagram
Summary
- 8‑bit unsigned range: 0 – 255; signed (two’s‑complement) range: –128 – +127.
- Binary addition follows simple sum‑and‑carry rules; always track the carry.
- Unsigned overflow = carry out of the MSB.
- Signed overflow = carry‑into MSB ≠ carry‑out MSB (or same‑sign addends give a result with opposite sign).
- Logical shifts provide rapid multiplication/division by powers of two; watch for overflow when shifting left.
- Conversion fluency (decimal ↔ binary ↔ hexadecimal) is essential for the IGCSE Computer Science exam.
Create an account or Login to take a Quiz
46 views
0 improvement suggestions
Log in to suggest improvements to this note.