arm - Assembly Language: How to stick a list of array into a word? -

how combine array of 4 byte 1 32-bit. first item should go significant nibble of result. store result in 32-bit variable result.

input: [list] = 0xc, 0x2, 0x6, 0x9 (each item byte, use dcb define variable of type byte)

output: [result] = 0x0c020609

edit answer:

add r1, r0 mov r1, r1, lsl #8 add r0, r0, #8 add r1, r0 mov r1, r1, lsl #8 add r0, r0, #8 add r1, r0 mov r1, r1, lsl #8 add r0, r0, #8 add r1, r0 

what you're describing same thing treating 4 contiguous bytes 32-bit integer stored in big-endian byte-order.

according gcc (on godbolt compiler explorer), best way byte-swap big-endian arm-native endian instruction arm provides explicitly purpose:

rev   r0, r0  

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#include <stdint.h> #include <endian.h> #include <string.h>  // type-punning unions alternative memcpy char array int union be_bytes {   uint32_t be_word;   char bytes[4]; };  uint32_t be_bytes_to_native( char *array ) {   union be_bytes tmp;   memcpy(tmp.bytes, array, 4);   // memcpy since take char* arg instead of union be_bytes * arg.   //  *think* (union be_bytes*)array safe, i'm not 100% sure.    // gnu c , many other compilers guarantee writing 1 union member , reading safe.  iso c doesn't, technically isn't portable.   return be32toh(tmp.be_word);   // endian.h, uses compiler builtins, inline asm, or c shift , mask instructions. } 

compiles to

be_bytes_to_native:     ldr     r0, [r0]  @ unaligned     rev     r0, r0     bx      lr 

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without rev instruction, @dwelch's answer on endianness conversion in arm suggests 4-instruction sequence byte-swapping 32-bit value in arm:

  eor r3,r1,r1, ror #16   bic r3,r3,#0x00ff0000   mov r0,r1,ror #8   eor r0,r0,r3, lsr #8 

note how combines use of barrel shifter instructions other mov. i'm still not sure add r0, r0, #8 (r0 += 8) in code supposed for.