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Comment by joosters | Hacker News Reader

So, I haven't written ARM assembly since I owned an Archimedes computer many years ago, but isn't there a mistake describing the very first assembly language instruction in lesson 1?

ldr r0,=0x20200000

Shouldn't the LDR be a MOV?

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LDR is a pseudo instruction that uses MOVs to load a register.

(http://www.woodmann.com/fravia/The%20ARM%20Processor.htm)

> The ARM assembler also supports a similar pseudo operation. The construct LDR rd,=value is used to load value into register rd. The LDR pseudo instruction uses the MOV or MOV instructions, or it places the constant in memory and uses program counter relative addressing to load the constant.

(http://sourceware.org/binutils/docs/as/ARM-Opcodes.html)

> If expression evaluates to a numeric constant then a MOV or MVN instruction will be used in place of the LDR instruction, if the constant can be generated by either of these instructions. Otherwise the constant will be placed into the nearest literal pool (if it not already there) and a PC relative LDR instruction will be generated.

Thanks! That makes sense, I'm only familiar with the old assembler from BBC Basic :) 0x202 wouldn't be a constant that you could use with MOV now I think about it, since it could only express integers generated with an 8 bit value and a shift.

ARM instructions are 32 bit, so there is a problem when you want to load a 32 bit literal. The assembler will get around this by sticking literals in the text section and then fixup the load with the correct offset.

Example, using a simple input:

  .globl _start
  _start:
    ldr r0, =0x20200000

Disassembles to:

  .globl _start
  _start:
    ldr r0, [pc, #-4]
    .word 0x20200000

Not a MOV and an ADD?

You can use that method (arm usually uses a "mov" and "movt" to load upper and lower half-words), but you would be using 2 instructions to accomplish the same thing that ldr with a literal pool does in one.

Not sure about a MOVT (my ARM assembler is only current to an ARM2...) but it's two (pipelined) instructions vs one instruction and a slow memory access. But then again, I'm thinking of a previous era with no instruction or data caches !

I speak mostly from an ARMv7 perspective (so pipelines, caches, prefetchers, etc are par for the course).

Your literal pool has a high degree of spacial locality WRT the instruction that references it. You can take measures to optimize this (see the ltorg directive) making I$ hits very likely.

In the end, the assembler is free to pick and choose which method is best for the target CPU and input code.

From what I can tell, yes. ldr r0, =0x20200000 loads from memory. In C, r0 = *0x20200000; what we would want is just r0 = 0x20200000; or mov r0, #0x20200000

They have it right, LDR acts differently if it's a register argument or register plus offset vs a immediate.

The immediate pseudo-instruction:

http://infocenter.arm.com/help/index.jsp?topic=/com.arm.doc....

The regular form:

http://infocenter.arm.com/help/index.jsp?topic=/com.arm.doc....

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