How can ARM's MOV instruction work with a large number as the second operand?

Question

I just begin to study ARM assembly language, and am not clear about how to use MOV to transfer an immediate number into a register.

From both the ARM reference manual and my textbook, it's said that range of immediate number following MOV instruction is 0-255. But when I test on my own PC in ADS 1.2 IDE, instruction

MOV     R2, #0xFFFFFFFF

performs well. Isn't number 0xFFFFFFFF out of range according to the specification?

Answer 1

A single ARM instruction can only encode an immediate constant that can be represented as an 8-bit immediate value, shifted by any even power of two.

However, there is also a MVN instruction, which is like MOV but inverts all the bits. So, while MOV R2, #0xFFFFFFFF cannot be encoded as a MOV instruction, it can be encoded as MVN R2, #0. The assembler may well perform this conversion for you.

Answer 2

Remember that the ARM can perform a certain set of manipulations on the immediate value as part of the barrel shifter that is incorporated into the ARM's opcodes.

This little article has one of the clearest explanations of some of the tricks that an ARM assembler can use to fit a large immediate number into the small available space of an ARM instruction:

• http://www.davespace.co.uk/arm/introduction-to-arm/immediates.html

The article discusses the trick likely used in your specific example of generating a MVN opcode to load the bitwise complement of the immediate value.

These kinds of manipulation can't be done with all immediate values, but the ARM assemblers are supposedly pretty smart about it (and C compilers certainly are). If no shift/complement tricks can be performed, the value will generally be loaded from a PC-relative location or maybe by 'building up' the value from several instructions.