Without the original opcodes, it's not possible to say with full confidence if this sequence is being accurately decoded.
However, one possibility which would be legitimate on a Cortex M3 would be this 32-bit Thumb-2 instruction as decoded by arm-none-eabi-objdump:
f803 1b01 strb.w r1, [r3], #1
This is of the type which ARM Holdings calls "Thumb syntax (wide, v6T2)" and an alternate formatting more closely matching your debugger's view would be:
STRB Rt, [Rn], #-255…255
In this case, the low byte of the value in r1 would be stored to the address in r3, and then r3 would be (as first pointed out by @old_timer) incremented by 1.
This is a rather plausible meaning, as the debug trace makes it appear that this is part of the implementation of the
memset() function, which fills memory with a byte value.
If this were the case, then the actual pseudo-code would be something like:
//r0=memset(r0, r1, r2)
Copy target address from r0 to working register r3
Add length in r2 to target address in r0 or r3 and save in r2
if r3 equals r2 goto end
store byte from r1 to address in r3, then increment r3 by one byte
return instruction, with the un-incremented target address in r0
As a further note, the range of increment for this instruction is +/- 255 units, where the unit in this case is a byte. A whole byte of the 32-bit instruction is available for storing the magnitude of the offset, while the sign is stored separately.
The copy of the base address from r0 to r3 is necessary because
memset() must return the unmodified base address - and conveniently, in the same r0 in which it was passed.