我正在尝试将输入的整数转换为二进制,我似乎无法在程序中找到错误。我可以加载输入的数字以及打印出一个8位二进制数字,但是没有一个正在打印。我究竟做错了什么?
enterNum:
.asciiz "Enter your number (base 10): "
printBaseTwo:
.asciiz "The number in base 2 is: "
#-----------------------------------------------------------------------
.text
#Print out string, collect intger input
main: li $v0, 4
la $a0, enterNum
syscall
li $v0, 5
syscall
move $t0, $v0
#create mask/print out the second string and prepare to print out binary
mask:
andi $t1, $zero, 1
sll $t1, $t1, 7
addi $t2, $zero, 8
li $v0, 4
la $a0, printBaseTwo
syscall
# compares mask to integer, starting at the most sig place
# if the mask is zero, print out zero
loop:
and $t3,$t0, $t1
beq $t3, $zero, print
add $t3, $zero, $zero
addi $t3, $zero, 1
j print
print:
#prepares to print integer in $a0
li $v0, 1
# moves either 1 or 0 into $a0 to be printed
move $a0, $t3
syscall
# shifts over right 1, getting closer to 0
srl $t1, $t1, 1
#lowers count
addi $t2, $t2, -1
#loop back to beginning if not finished printing binary Num
bne $t2, $zero, loop
beq $t2, $zero, exit
exit:
li $v0, 10
syscall
答案 0 :(得分:1)
我从32位MIPS版本开始回答这个问题。
首先,当要求十进制数时,MIPS将其保存为寄存器中的32位数字。因此,我做了以下调整:
andi $t1, $zero, 1
sll $t1, $t1, 7
addi $t2, $zero, 8
变为:
add $t1, $zero, 1
sll $t1, $t1, 31
addi $t2, $zero, 32
请注意我如何将andi
更改为addi
。如果您使用的是8位版本的MIPS,那么这将是唯一必要的调整。
我希望我解决了你的问题,它可以在MARS模拟器中工作!