Question

我发现这段代码片段我认为会将整数转换为十六进制。但是，我根本不关注它。我添加了评论说我认为发生了什么，但我不知道为什么要这样做。所以，假设我正确地注意到每条线路正在做什么，有人可以向我解释为什么要这样做吗？如何以任何方式帮助转换为十六进制？

$ a0是整数值

$ a1是结果应该在哪里的地址

        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使用分支延迟时隙，这意味着分支指令之后的指令总是在分支被采用（或不被采用）之前执行。

所以这就是说“如果$ t0小于10（即在0..9范围内），转到DIGIT，但不管$ t0的值如何，首先加48（ASCII'0'）。分支被取出你现在已经从0..9转换为'0'..'9'。如果没有采取分支，$ t0最初在10..15范围内，现在将在范围58..63，所以我们再添加39个以获得97..102范围内的值（'a'...'f'的ASCII代码）“。

MIPS汇编：从整数转换为十六进制

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