movw and movt in arm assembly

Question

I'm having trouble deciphering this block of assembly code. What would the value of r1 be by the end and how would I get there?

3242ba66    f6454118    movw    r1, 0x5c18
3242ba6a        466f    mov     r7, sp
3242ba6c    f6c0415a    movt    r1, 0xc5a
3242ba70    f2460002    movw    r0, 0x6002
3242ba74    f6c0405a    movt    r0, 0xc5a
3242ba78        4479    add     r1, pc
3242ba7a        4478    add     r0, pc
3242ba7c        6809    ldr     r1, [r1, #0]

Answer 1

movw followed by a movt is a common way to load a 32-bit value into a register. It's the equivalent of OR-ing those two immediate values together, with the movt being the upper 16-bit. In this case, r1 = (movt immediate value << 16) | (movw immediate value)).

3242ba66    f6454118    movw    r1, 0x5c18   // r1 = 0x5c18
3242ba6a        466f    mov     r7, sp
3242ba6c    f6c0415a    movt    r1, 0xc5a    // r1 = (r1 & 0xffff) | (0xc5a << 16)
3242ba70    f2460002    movw    r0, 0x6002
3242ba74    f6c0405a    movt    r0, 0xc5a
3242ba78        4479    add     r1, pc       // r1 = r1 + pc
3242ba7a        4478    add     r0, pc
3242ba7c        6809    ldr     r1, [r1, #0] // r1 = *(r1 + 0)