Foundations — Instruction formats and encoding

This page assumes you have seen nothing. We will build every symbol the parent note Instruction formats and encoding uses — one at a time, each one anchored to a picture, each one earned before we use it. Nothing here is memorised; everything is derived.

0. The most basic object: a bit

Figure s01 below draws a row of eight bulbs: lit bulbs (butter-yellow) are 1s, dim grey bulbs are 0s. Look at the leftmost bulb — it is one bit, and the little arrow points to it. The green line underneath brackets all eight bulbs together; that grouping is what we will call a byte in a moment.

1. A group of bits: binary numbers and $2^b$

If one bit holds 2 possibilities, what does a group of $b$ bits hold?

Add one bit → each old pattern can now end in 0 or 1 → the number of patterns doubles. So:

• 1 bit → 2 patterns
• 2 bits → 4 patterns
• 3 bits → 8 patterns
• $b$ bits → $2^b$ patterns.

Figure s02 turns this into a bar chart: each bar is taller than the one before it by a factor of two, and the coral arrows between bars say "×2 again". That visual doubling is the meaning of $2^b$.

From a bit pattern to an actual number: positional weights

We keep saying "$10110$ is a number" — let us earn that, because the whole topic leans on it.

Take $10110$ (five bits). Line up positions from the right:

$\underset{2^4=16}{1}\underset{2^3=8}{0}\underset{2^2=4}{1}\underset{2^1=2}{1}\underset{2^0=1}{0}$

Add the weights where the bit is $1$: $16+4+2=22$.

2. The inverse question: ${\log }_2$ and the ceiling $\lceil \rceil$

The parent note writes $b=\lceil {\log }_{2}N\rceil$. Let us earn every symbol.

The question we are really asking: "I have $N$ different things (say 32 registers). How many bits $b$ do I need so that $2^b$ is at least $N$?"

But ${\log }_{2}100=6.64\dots$ — you cannot have 6.64 switches. Bits are whole, and 6 bits ($=64$ patterns) is too few. We must round up to 7.

Figure s03 puts $2^5=32$, $2^6=64$, $2^7=128$ on a line and drops a coral marker at "need 100". You can see 64 sits left of the marker (too few) and 128 sits right (fits), so the answer must round up from $6.64$ to $7$.

3. A named group of bits: a field

Before we name fields we must agree how to count the bit positions — otherwise "bits 0 through 6" is ambiguous.

Figure s04 draws the 32-bit word as six rounded, pastel boxes. Notice the labels along the bottom: "high bits (left)" and "bit 0 (right)" — that is our convention made visible. The coral opcode box sits at the far right (bits 0–6); the arrow reminds you the decoder looks there first.

The two field-names the parent leans on:

• opcode — the "what job" field (add? jump? load?). Always in the same position so the decoder finds it first.
• operand fields — the "what to do it to": register numbers, or an immediate.

4. Two special operands: register and immediate

$32$ registers $\Rightarrow b=\lceil {\log }_{2}32\rceil =5$ bits per register field. That "5" you keep seeing is not magic — it is Section 2 applied to 32.

5. Negative immediates: two's complement and sign extension

addi x1, x2, -8 must work, so a field must be able to mean a negative number.

Building it from the positional weights of Section 1. In an unsigned $b$-bit number every bit has weight $+2^i$. Two's complement changes exactly one thing: the top bit's weight becomes negative, $-2^{b-1}$, while every lower bit keeps its usual $+2^i$.

Watch it work on 4 bits ($b=4$, so the top weight is $-2^3=-8$):

$1000=-8,1111=-8+4+2+1=-1,0111=4+2+1=+7.$

The range then falls straight out of the weights. The most negative value turns on only the top bit: $1000\dots 0=-2^{b-1}$. The most positive turns on every bit except the top: $0111\dots 1=2^{b-1}-1$. So:

$\text{range}=-2^{b-1}\text{to}{2}^{b-1}-1.$

For the 12-bit immediate the parent quotes: $-2^{11}$ to $2^{11}-1=-2048$ to $+2047$. See Two's Complement and Sign Extension for the full treatment.

6. Putting the symbols together

Now every symbol in the parent's tally makes sense on sight:

$\underset{7}{\underbrace{funct7}}\underset{5}{\underbrace{rs2}}\underset{5}{\underbrace{rs1}}\underset{3}{\underbrace{funct3}}\underset{5}{\underbrace{rd}}\underset{7}{\underbrace{opcode}}=32\text{ bits}$

• $7,5,5,3,5,7$ are field widths in bits (Section 3).
• Each $5$ came from $\lceil {\log }_{2}32\rceil$ (Sections 2 & 4).
• The sum being exactly $32$ is the fixed budget the parent keeps stressing.

Related building topics you can now approach: Addressing Modes (how operand fields name memory), RISC vs CISC (fixed vs variable width philosophy), Pipelining (why fast fixed decode matters).

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