编写一个小子程序，将0到15之间的数字转换为可打印的ASCII编码字符："0"到"9"，或"A"到"F"，具体取决于数字。对于不在0到15范围内的数字，一些位将被忽略。
名称：子例程必须称为hexasc。参数：一，在寄存器$a0中。4个最低有效位指定一个数字，从0到15。寄存器$a0中的所有其他位可以有任何值，必须忽略。返回值：寄存器$v0中的7个最低有效位必须是ASCII码，如下所述。当函数返回时，所有其他位必须为零。必需操作：函数必须将0到9的输入值转换为数字的ASCII代码分别为"0"到"9"。输入值10到15必须转换为字母"A"到"F"。
.text
main:
li $a0,0 # change this to test different values
jal hexasc # call hexasc
nop # delay slot filler (just in case)
move $a0,$v0 # copy return value to argument register
li $v0,11 # syscall with v0 = 11 will print out
syscall # one byte from a0 to the Run I/O window
stop: j stop # stop after one run
nop # delay slot filler (just in case)
我在这里写了我的hexasc子程序，由于某种原因，我不理解，所以它不起作用。我不知道我在哪一部分有错误，也许我没有按照上面规范中的Say做。如有任何帮助，我们将不胜感激。
hexasc:
addi $sp,$sp,-4 #make space on the stack
sw $a0,0($sp) #store $a0 on the stack
li $t0,0x30 #$t0 = 0x30 ('0' in ascii)
andi $a0,$a0,0xf #only 4 least significant bits is
#needed ignore other bits
loop:
add $a0,$a0,$t0 #$a0 i will think why i did this
addi $t0,$t0,1 #increment $t0 by 1
beq $t0,0x39,loop2 # if $t0 = 0x39 (9 in ascii)
j loop
li $t1,0x41 # $t1 = 0x41 ( A in ascii)
loop2:
andi $a0,$a0,0xf # only 4 LSB ignore other bits
add $a0,$a0,$t1 # ???? i will think about this
beq $t1,0x46,done # if $t1 = 0x46 (F in ascii)
j loop2
done:
add $v0,$a0,$a0 # return $a0 in $v0 suspicious ...?
lw $a0,0($sp) # restore the $a0
addi $sp,$sp,4 # put back the stack
jr $ra
这两篇文章中有一个小错误值得一提，li $v0,0x41应该是li $v0,0x7。这是因为您指定10的输入值应该以ASCII打印字符A，但如果您有li $v0,0x41，则10的输入值的输出将是K而不是A。这是hexasc的另一个版本，它可以打印出你规范中的Say’s。
hexasc:
andi $a0,$a0,0x0f
addi $v0,$zero,0x30
addi $t0,$zero,0x9
ble $a0,$t0,L1
nop
addi $v0,$v0,0x7
L1:
add $v0,$a0,$v0
jr $ra
nop
一些问题。。。
您在十进制测试期间销毁了$a0值，因此它的十六进制部分的值错误。
你的跳跃是在做无限循环。他们应该去done:
我已经创建了您代码的两个版本。一个带有注释，一个带有更正的代码[请原谅免费的样式清理]。
这是注释版本：
hexasc:
# NOTE/BUG: stack save/restore of $a0 is not needed due to MIPS ABI
addi $sp,$sp,-4 # make space on the stack
sw $a0,0($sp) # store $a0 on the stack
li $t0,0x30 # $t0 = 0x30 ('0' in ascii)
andi $a0,$a0,0xf # only 4 least significant bits is
# needed ignore other bits
# NOTE/BUG: altering $a0 further destroys the value for the hex test
loop:
add $a0,$a0,$t0 # $a0 i will think why i did this
addi $t0,$t0,1 # increment $t0 by 1
# NOTE/BUG: beq is wrong -- we want bgt
# NOTE/BUG: the range test for decimal should be 0x09 and _not_ 0x39
beq $t0,0x39,loop2 # if $t0 = 0x39 (9 in ascii)
# NOTE/BUG: looping is wrong -- this should be 'j done'
j loop
li $t1,0x41 # $t1 = 0x41 ( A in ascii)
# NOTE/BUG: $a0 is now wrong because of add above
loop2:
andi $a0,$a0,0xf # only 4 LSB ignore other bits
add $a0,$a0,$t1 # ???? i will think about this
# NOTE/BUG: no range check needed as all values are in range
beq $t1,0x46,done # if $t1 = 0x46 (F in ascii)
# NOTE/BUG: no looping required
j loop2
done:
add $v0,$a0,$a0 # return $a0 in $v0 suspicious ...?
lw $a0,0($sp) # restore the $a0
addi $sp,$sp,4 # put back the stack
jr $ra
以下是更正后的版本：
hexasc:
andi $a0,$a0,0xf # only 4 least significant bits is needed
# check for decimal
li $v0,0x30 # set 0x30 ('0' in ascii)
ble $a0,0x09,hexasc_done # value in range 0x0-0x9? if yes, fly
li $v0,0x41 # set 0x41 ('A' in ascii)
hexasc_done:
add $v0,$v0,$a0 # convert value to ascii
jr $ra