Review of knowledge for understanding the data movement instructions

Review 1: Instructions are stored in RAM (memory)

Review of knowledge for understanding the data movement instructions

Review 2: The CPU fetches instructions into its Instruction Register first:

The fetched instruction is then executed (by steps 2,3,4 of the instruction execution cycle)

The data "movement" instructions in ARM

Important fact about assembler programming:

all computations begin by moving values into registers and
end with storing the result into memory

We will study the data transfer instructions:

The mov (move) instruction:

to "move" (= copy) a small (binary) number (< 100) into a register and
to "move" (= copy) the value (= binary number) from one register into another register

The movw and movt instructions to "move" (= copy) large numbers into a register
The ldr (load) instructions to fetch a (binary) number in memory into a register
The str (store) instructions to store a (binary) number in a register into RAM (memory)

Assigning a small constant to a register using the mov instruction

The first way to use the mov instruction:

The mov instruction can be used to assign a small (integer) constant to a register

The syntax of this mov instruction is:

mov rN, #constant // rN = any ARM register name Examples: mov r0, #0 // Assign the value 0 to register r0 mov r2, #-1 // Assign the value -1 to register r2 mov pc, #4 // Assign the value 4 to the Program Counter

The constant will be stored as a 32 bit 2's complement number in the register rN

Moving a small (constant) number into a register - example

Examples:

mov r0, #15 // Stores 00000000000000000000000000001111 in r0 mov r1, #-15 // Stores 11111111111111111111111111110001 in r1

Effect of the mov instructions:

DEMO: home/cs255001/demo/asm/2-mov/mov-intro.s

Also try: mov r0, #123456 (1e240₍₁₆₎ = 123456₁₀)

How the computer executes the mov r0,#15 instruction

Suppose the next instruction executed is mov r0, #15:

The mov r0, #15 instruction is first fetched into the Instruction Register (IR) by the Instruction Fetch step

How the computer executes the mov r0,#15 instruction

Then the mov r0, #15 will be executed by steps 2,3,4 of the Instruction Execution Cylce:

The mov r0, #15 will store the (binary) number 15 (= 00..1111) into register R0

Why can the mov instruction only use a small constant ?

The mov instruction is encoded as follows:

mov rN, #x is encoded in ARM as: +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | | | | |0|0|1|1|1|0|1|S|x|x|x|x| N | value (x) | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ <-----> ^ <-----> <-----> <---------------------> 1 1 1 0 | "mov" not used 12 bits unconditional | immediate bit (= use x as a number)

Notice: there are only 12 bits available in the instruction code to encode the value x !!!

Therefore:

the mov instruction can only use binary numbers that can be represented by at most 12 bits

The range of values that can be used in the mov rN, #x

Simplified rule for using mov rN, #x:

mov rN, #x can correctly use values in the range x ∈ [-100, 100]

Comment:

the exactly "range" of values is very complicated
Also, you do not to this knowledge to learn assembler programming

For the curious: you can read the online lecture note for the explanation: click here )

What if I need to use a large number into a register ?

Important fact:

The mov rN, #x instruction cannot use large values
Example: this mov instruction will cause a assembler/compile error:
mov r0, #1234567 // Compile error

How to move a large integer value(= 32 bits) into a register in ARM assembler:

Divide the 32 bits number into two (2) 16 bits numbers
Use the movw instruction to ove the lower half into the register
Use the movt instruction to ove the upper half into the register

We will learn how to do this soon