Relocation in assembly

Question

I have a boot-up code for a bare-metal ARM written in assembly and I'm trying to understand how it works. The binary is written in some external Flash, and is copying parts of itself in RAM at boot-up. I still didn't exactly get the concept of relocation in this context, even though I read this wikipedia entry. The RAM is mapped to a low address window, and the flash in a high address window. Can someone explain to me why we test the value of the link register here?

/* Test if we are running from an address, we are not linked at */
       bl check_position
 check_position:
        mov     r0, lr                  
        ldr     r1, =check_position
        cmp     r0, r1                  /* ; don't relocate during debug */
        beq     relocated_entry 

Answer 1

My guess is the application runs from ram, and when debugging the application this author is perhaps using some sort of bootloader and or jtag to load the test app directly into ram, thus no reason to copy and run (which could cause a crash).

Another reason you would do something like this is to avoid an infinite loop. If for example you want to boot from flash (have to usually) but execute from ram, the simplest way to do that is to just copy the whole flash or whole some chunk of flash to ram and just branch to the start of ram. Which when you do that means you hit the "copy the app to ram and branch" loop again, to avoid it the second time (which might crash you), you have some sort of am I running this loop from flash or not test.