Hello. I've been looking at a piece of code that I'm not entirely sure what it does.
LDR r0, =str MOV r1, #0 loop LDRB r2, [r0], #1 ADD r1, r1, #1 CMP r2, #0 BNE loop SUB r1, r1, #1 Data Segment Little Endian Format str DCB "mCQ5Gtq",0
From my understanding, DCB loads the string into a word of 32 bits, and each character has its own byte. Little Endian Format means the least significant byte is stored in the lower memory address. LDR makes r0 = the string. LDRB loads the least significant byte held in r0, into r2 offset by 1 bit.
1) Does the least significant byte start with m or q? i.e. would it be <- 0000 0017 (ascii for q) or <- 0000 006D (ascii for m).
2) Does the LDRB operator load the least significant byte from r0 offset by 1 bit, into r2. If so, what would r2 be (just the first loop).
Thank you very much
Equivalent to
char str[] = "mCQ5Gtq"; uint8_t *r0 = str; // pointer is 32-bit wide uint32_t r1 = 0; do { r2 = *r0++; // LDRB r2, [r0], #1 r1++; }while(r2); // while not NUL r1--;
R0 is a 32-bit pointer, for a byte, it reads the byte into R2, and increments R0 by 1
You could always run it in the Simulator - and watch what it does ... ?