MIPS汇编：从整数转换为十六进制

Question

我发现这个代码片断，我相信将整数转换为十六进制。但是，我没有遵循它。我添加了说明我相信正在发生的意见，但我不知道为什么这样做。所以，假设我正确地注意到每一行都在做什么，请有人向我解释为什么这样做？至于它如何以任何方式帮助转换为十六进制？

$ A0为整数值

$ A1是它的结果应该是

 addi $t0, $0, 48  #set $t0 equal to 48 
     sb $t0, 0($a1)   #store $to (48) at location 0 in $a1 
     addi $t0, $0, 120  #set $t0 equal to 120 
     sb $t0, 1($a1)   #store $t0 (120) at location 1 in $a1 
     addi $t1, $a1, 9  #set $t1 = the address + 9 

LOOP: 

     andi $t0, $a0, 0xf #$t0 = 1 if $a0 and 0xf are the same (0xf = beginning of hex)? 

     slti $t2, $t0, 10  #if $t0 is less than 10, $t2 = 1, else 0 
     bne $t2, $0, DIGIT #if $t2 does not equal 0, branch to DIGIT 
     addi $t0, $t0, 48  #set $t0 equal to 48 
     addi $t0, $t0, 39  #set $t0 equal to 39 (why did we just write over the 48?) 
DIGIT: 

     sb $t0, 0($t1)  #set $t0 equal to whatever's in location 0 of $t1 

     srl $a0, $a0, 4  #shift right 4 bits 

     bne $a0, $0, LOOP  #if $a0 does not equal 0, branch to LOOP 
     addi $t1, $t1, -1  #set $t1 = $t1 - 1 

DONE: 

     jr $ra    #set the jump register back to $ra 
     nop 

Answer 1

slti $t2, $t0, 10  #if $t0 is less than 10, $t2 = 1, else 0 
    bne $t2, $0, DIGIT #if $t2 does not equal 0, branch to DIGIT 
    addi $t0, $t0, 48  #set $t0 equal to 48 
    addi $t0, $t0, 39  #set $t0 equal to 39 (why did we just write over the 48?) 

MIPS的地址使用分支延迟槽，这意味着始终遵循分支指令的指令在分支采取（或未采取）之前执行。

所以这说的是“如果$ t0小于10（即在0..9范围内），转到DIGIT，但首先加48（ASCII'0'），无论$ t0的值如何。如果分支被采用，你现在已经从0..9转换为'0'..'9'。如果分支未被采用，$ t0原来在10..15的范围内，现在将在58..63的范围内，所以我们再添加39个以获得范围在97..102（'a'..'f'）的ASCII码的值“。