Tape Writer

The least significant byte of the register is written to memory first because it is already where the sb instruction needs it. Then the remaining bytes of $9 are shifted into the right-most byte one by one and written to memory. Here is a complete version of the program:

## endian.asm
##
## copy $9 to memory in big-endian order
##
## Register Use:
## $8  --- first byte of the tape buffer
## $9  --- 4-byte integer

      .text
      .globl  main

main: 
      lui  $9,0x1234      # put data in $9
      ori  $9,0x5678      #
      lui  $8,0x1000      # $8 is base register
      sb   $9,3($8)       # least significant byte
      srl  $9,$9,8        # move next byte to low order
      sb   $9,2($8)       # bits 8-15 
      srl  $9,$9,8        # move next byte to low order
      sb   $9,1($8)       # bits 16-23 
      srl  $9,$9,8        # move next byte to low order
      sb   $9,0($8)       # most significant byte

      .data
tape:                     # base register points here
      .space 1024         # tape buffer (1K bytes)
         

The .space directive reserves bytes in memory, in this case 1024₁₀ bytes. Pretend this is the buffer from which a tape record will be written. The example program deals with just the first four bytes, but in a real program all 1K bytes of the buffer would be collected before the buffer was written to tape.