Memory and addresses.

Step 1 of 5

Sixteen cells

Before a computer can work anything out, it needs somewhere to keep numbers. That place is the memory: a row of cells, each holding one number. To find a cell, each one gets a number of its own, its address, like house numbers on a street.

In the Scene a toy processor sits on a board. Look inside is open at Blocks: the lid is lifted, and to the right of the chip you can see the memory, sixteen cells with their addresses. The processor picks a cell by sending its address out on the address bus, a ribbon of parallel wires. A lit lane is a wire at high voltage, 1; a dark lane is 0. Each wire carries one bit.

Step 2 of 5

One byte

Every memory cell is a byte: eight bits. Read as a number in base two, each bit has a place value, right to left: $1, 2, 4, 8, 16, 32, 64, 128$⁠. Add up the place values of the bits that are on, and you have the number. 0001 1100, for example, is $16 + 8 + 4 = 28$⁠, the number in cell 14.

The registers inside the processor, A and B, are eight bits wide too, so they also hold one byte each.

Step 3 of 5

A byte is a number

The memory does not hold only data. The program sits there too: the first five cells, addresses 0 to 4, are the instructions the processor will run, and the last two cells are the numbers the program adds. Nothing in a byte marks it as an instruction or a number. The labels on the memory are for us; the chip only ever sees bits.

Now the lid is closed. The byte at address 0 is 0001 1110. Read it as a number.

Step 4 of 5

Opcode and address

When the same byte is read as an instruction, it splits in two. The left four bits are the opcode: what to do. The right four are an address: which cell to do it to. Four bits are just enough for sixteen cells.

This chip has eight opcodes:

Code	Instruction	What it does
0001	LDA	loads register A with what is in the cell
0010	LDB	loads register B with what is in the cell
0011	ADD	adds A and B, and the answer goes into A
0100	SUB	takes B from A, and the answer goes into A
0101	STA	writes A to the cell
0110	JMP	jumps to the address
0111	JZ	jumps to the address if the last answer was zero
1111	HLT	stops the processor

So 0001 1110 is LDA 14: opcode 0001, load A, and address 1110, that is 14. The program in the Scene is LDA 14, LDB 15, ADD, STA 13, JMP 4: load the two numbers, add them, store the sum in cell 13, then jump on the spot for ever.

Step 5 of 5

Copy, don’t move

Clock tick stands at 1. The instruction LDA 14 is already in the instruction register, and on the next tick it will be carried out: address 14 goes out to memory, and the number in that cell comes to register A. Predict what will be left in cell 14 afterwards.