4 ms·

so... what is the proper way to set a register to the value 0x101, or any other value that cannot be represented?

by webkike 9y ago

JoshTriplett 9y ago
Use two instructions; load part of it then load the rest.
- gpvos 9y ago
  If I understand correctly, for some values that would take four instructions.
  - jhawthorn 9y ago
    You would use the LDR instruction to load the word from RAM. So just one instruction, but with some slower memory access. When writing ARM assembly, the assembler (at least the ones I've seen) abstract this detail away from you. You can use ldr r0, =0xff000000 and the assembler will emit a MOV if it can, and if it can't it will put the value somewhere in RAM and use LDR to load it up. ARM is fun.
  - hrydgard 9y ago
    Modern ARM actually has a 16-bit immediate version of MOV, plus MOVT which is the same but moves to the top 16 bits of a register. With a pair of those, you can load any 32-bit value in 2 instructions, without polluting the data cache like you would with LDR [pc+offset].
deleted 9y ago
[deleted]
brandmeyer 9y ago
In ARMv7 and previous, the program counter is an ordinary register, so you can perform a load relative to the PC for anything that doesn't fit into a register's literal field. Compilers typically emit so-called literal pools before the entry or after the exit point of each function (or sometimes before a basic block inside very large functions). Some compilers support literal pool merging as a code size reduction technique, too. Edit: Also, ARMv8 has dedicated instructions for for pc-relative loads and jumps.
repiret 9y ago
On ARM1156, and ARMv7 and newer, use MOVW, which takes a simple 16-bit immediate. If 16 bits doesn’t do it, use a MOVW/MOVT pair which can load any 32-bit value. On older ARMs, either do a load immediate followed by an immediate arithmetic, like so: MOV R0, 0x0100 ADD R0, R0, 0x01 Or use a PC relative load: LDR R0, [PC, offset where you stored 0x101] Which the assembler allows you to abbreviate as: LDR R0, =0x101 And it will find a spot for 0x101 and generate the correct PC relative load.

so... what is the proper way to set a register to the value 0x101, or any other value that cannot be represented?

JoshTriplett 9y ago

Use two instructions; load part of it then load the rest.

gpvos 9y ago
If I understand correctly, for some values that would take four instructions.
- jhawthorn 9y ago
  You would use the LDR instruction to load the word from RAM. So just one instruction, but with some slower memory access. When writing ARM assembly, the assembler (at least the ones I've seen) abstract this detail away from you. You can use ldr r0, =0xff000000 and the assembler will emit a MOV if it can, and if it can't it will put the value somewhere in RAM and use LDR to load it up. ARM is fun.
- hrydgard 9y ago
  Modern ARM actually has a 16-bit immediate version of MOV, plus MOVT which is the same but moves to the top 16 bits of a register. With a pair of those, you can load any 32-bit value in 2 instructions, without polluting the data cache like you would with LDR [pc+offset].

deleted 9y ago

[deleted]

brandmeyer 9y ago

In ARMv7 and previous, the program counter is an ordinary register, so you can perform a load relative to the PC for anything that doesn't fit into a register's literal field. Compilers typically emit so-called literal pools before the entry or after the exit point of each function (or sometimes before a basic block inside very large functions). Some compilers support literal pool merging as a code size reduction technique, too. Edit: Also, ARMv8 has dedicated instructions for for pc-relative loads and jumps.

repiret 9y ago

On ARM1156, and ARMv7 and newer, use MOVW, which takes a simple 16-bit immediate. If 16 bits doesn’t do it, use a MOVW/MOVT pair which can load any 32-bit value. On older ARMs, either do a load immediate followed by an immediate arithmetic, like so: MOV R0, 0x0100 ADD R0, R0, 0x01 Or use a PC relative load: LDR R0, [PC, offset where you stored 0x101] Which the assembler allows you to abbreviate as: LDR R0, =0x101 And it will find a spot for 0x101 and generate the correct PC relative load.