What does the hash (#) value associated with the ARM LDR instruction mean?

Question

I'm trying to debug a crash I am experiencing in my application. The stack trace is pointing to an LDR instruction with the following format (thanks disassembler):

LDR R3, [R0,#4]

My question is in regards to the source component. What does the #4 in the second parameter mean? I'm assuming it is some kind of offset, but I haven't found documentation supporting that for the LDR instruction.

In general in arm assembler the # is used to indicate an immediate value. — old_timer, May 23 '11 at 15:24

score 12 · Accepted Answer · answered May 23 '11 at 13:38

12

It loads R3 from the address in R0 + 4 bytes. So, yes, it is a byte offset. See this explanation of the addressing modes.

answered May 23 '11 at 13:38

Graham Borland

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For a really comprehensive coverage, always wise to read [ARM Assembly Language by Pete Cockerell](http://www.peter-cockerell.net/aalp/html/frames.html) [(PDF)](http://www.peter-cockerell.net/aalp/resources/pdf/all.pdf) - a good book, quite old, but still possibly the best ARM learning aid. – Orbling May 23 '11 at 13:53

score 2 · Answer 2 · answered Sep 23 '14 at 07:04

2

It adds 4 to the value in R0 and uses that as the address to load a 32 bit value into the register R3

answered Sep 23 '14 at 07:04

Yatin Prayag

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score 2 · Answer 3 · answered Aug 23 '18 at 14:24

In GNU gas, the hash # is only required for ARMv7 when not using .syntax unified

For example, you can write it without # for ARMv8 aarch64-linux-gnu-as:

LDR x0, [x0,4]

or if use .syntax unified in arm-linux-gnueabihf-as:

.syntax unified
LDR x0, [x0,4]

More details at: Is the hash required for immediate values in ARM assembly?

What does the hash (#) value associated with the ARM LDR instruction mean?

3 Answers3

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