Add together two positive binary patterns, apply logical and arithmetic binary shifts, and understand the concept of overflow in relation to the number of bits available to store a value.

What this dot point is asking

Edexcel wants you to add two positive binary numbers, carry out logical and arithmetic binary shifts and state their effect, and explain overflow: what it is, when it happens, and why the stored result is then wrong.

Binary addition

Edexcel asks you to add two positive binary patterns, so each column totals at most 3 (two bits plus a carry), and the carry is never more than 1. The processor does this in its arithmetic logic unit: a full adder adds two bits plus a carry-in and produces a sum bit and a carry-out, which feeds the next column, which is why addition runs right to left.

Binary shifts

Shifting is much faster for the CPU than full multiplication or division, which is why it is used to multiply or divide by powers of 2. For a logical shift, always fill with 0. Examples: $00000110$ (6) shifted left by 1 gives $00001100$ (12, that is $6 \times 2$); $00010100$ (20) shifted right by 2 gives $00000101$ (5, that is $20 \div 4$). Bits pushed off the right in a right shift are discarded, so precision can be lost, and bits pushed off the left in a left shift can cause overflow.

An arithmetic right shift differs only in the fill: it copies the most significant (sign) bit into the vacated left positions, so a negative two's complement number stays negative. This keeps division by 2 correct for signed numbers.

Overflow

For an unsigned 8-bit register the largest value is $11111111 = 255$, so any addition whose true total exceeds 255 overflows. For example $11111111 + 00000001$ should be 256, which needs the 9 bits $100000000$; the leading 1 is lost and the register holds $00000000$, the wrong answer. Overflow also strikes shifts: a left shift that pushes a 1 off the left end loses it, so the multiply-by-2 result is wrong. The fixed number of available bits is the root cause, so a question about overflow always comes back to "the result needs more bits than the register holds".

Try this

Q1. Add the 8-bit binary numbers $00001111$ and $00000001$. [2 marks]

• Cue. $00010000$ ($15 + 1 = 16$).

Q2. State the effect on the denary value of a logical right shift of 1. [1 mark]

• Cue. It divides the value by 2 (with any bit shifted off the right discarded).