Why use LDR over MOV (or vice versa) in ARM assembly?
I'm looking through this tutorial: http://www.cl.cam.ac.uk/freshers/raspberrypi/tutorials/os/ok01.html
The first line of assembly is:
ldr r0,=0x20200000
the second is:
mov r1,#1
I thought ldr was for loading values from memory into registers. But it seems the = means the 0x20200000 is a value not a memory address. Both lines seem to be loading the absolute values.
Answer
It is a trick/shortcut. say for example
ldr r0,=main
what would happen is the assembler would allocate a data word, near the instruction but outside the instruction path
ldr r0,main_addr
...
b somewhere
main_addr: .data main
Now expand that trick to constants/immediates, esp those that cannot fit into a move immediate instruction:
top:
add r1,r2,r3
ldr r0,=0x12345678
eor r1,r2,r3
eor r1,r2,r3
b top
assemble then disassemble
00000000 <top>:
0: e0821003 add r1, r2, r3
4: e59f0008 ldr r0, [pc, #8] ; 14 <top+0x14>
8: e0221003 eor r1, r2, r3
c: e0221003 eor r1, r2, r3
10: eafffffa b 0 <top>
14: 12345678 eorsne r5, r4, #125829120 ; 0x7800000
and you see the assembler has added the data word for you and changed the ldr into a pc relative for you.
now if you use an immediate that does fit in a mov instruction, then depending on the assembler perhaps, certainly with the gnu as I am using, it turned it into a mov for me
top:
add r1,r2,r3
ldr r0,=0x12345678
ldr r5,=1
mov r6,#1
eor r1,r2,r3
eor r1,r2,r3
b top
00000000 <top>:
0: e0821003 add r1, r2, r3
4: e59f0010 ldr r0, [pc, #16] ; 1c <top+0x1c>
8: e3a05001 mov r5, #1
c: e3a06001 mov r6, #1
10: e0221003 eor r1, r2, r3
14: e0221003 eor r1, r2, r3
18: eafffff8 b 0 <top>
1c: 12345678 eorsne r5, r4, #125829120 ; 0x7800000
So it is basically a typing shortcut, understand that you are giving the assembler the power to find a place to stick the constant, which it usually does a good job, sometimes complains, not sure if I have seen it fail to do it safely. Sometimes you need a .ltorg or .pool in the code to encourage the assembler to find a place.