为什么我的MIPS基本转换器在当前值之后打印出先前循环中的值? [英] Why is my MIPS base converter printing out the values from a previous loop after the current values?
问题描述
我是MIPS的新手，这让我感到完全困惑.我制作了一个程序来转换基数，并且在第一次运行时工作良好，但是当它循环时，它显示了循环先前迭代中其他寄存器的值.输出如下.我已经尝试了所有我能想到的东西，但我没有主意...
I'm very new to MIPS and this has me completely baffled. I made a program to convert bases and it works fine the first time through, but when it loops, its displaying the values from the other registers from the previous iterations of the loop. The output is below. I've tried everything that I can think of and I'm out of ideas...
Enter a decimal number: 10
The number in base 2 is 00000000000000000000000000001010
The number in base 4 is 0000000000000022
The number in base 16 is 0000000A
The number in base 8 is 0000000012
您想输入另一个号码吗?1
Would you like to input another number? 1
Enter a decimal number: 11
The number in base 2 is 0000000000000000000000000000101100000000000000220000000A0000000012
The number in base 4 is 00000000000000230000000A0000000012
The number in base 16 is 0000000B0000000012
The number in base 8 is 0000000013
您想输入另一个号码吗?
Would you like to input another number?
.text
.globl __start
__start:
la $a0, prompt # Prompt for a base10 integer
li $v0, 4
syscall
li $v0, 5
syscall
move $a0, $v0
jal bin
jal base4
jal hex
jal base8
la $a0, endl
li $v0, 4
syscall
la $a0, repeat
li $v0, 4
syscall
li $v0, 5
syscall
beqz $v0, eop
la $a0, endl
li $v0, 4
syscall
j __start
eop:
li $v0,10 # End Of Program
syscall
##########################################################
#
# BASE 16
#
##########################################################
hex:
sub $sp, $sp, 24 # Push register onto stack
sw $a0, 0($sp)
sw $s0, 4($sp)
sw $s1, 8($sp)
sw $s2, 12($sp)
sw $s3, 16($sp)
sw $s4, 20($sp)
move $s2, $a0 # Move a0 to s2
la $a0, ans3 # Display string before hex answer
li $v0, 4
syscall
li $s0, 8 # 8 digits for hex word
la $s3, hexresult # Hex string set up here
hexloop:
rol $s2, $s2, 4 # Start with leftmost digit
and $s1, $s2, 0xf # Mask 15 digits in s2 and place results in s1
ble $s1, 9, hexprint # If s1 <= 9, go to print
add $s1, $s1, 7 # Else s1 = s1 + 7 (to get A-F)
hexprint:
add $s1, $s1, 48 # Add 48 (30 hex) to get ascii code
sb $s1,($s3) # Store byte in result. s3 -> result
add $s3, $s3, 1 # s3 = s3 + 1
add $s0, $s0, -1 # s0 = s0 - 1
bnez $s0, hexloop # If s0 != 0, go to hexloop
la $a0, hexresult # display result
li $v0, 4
syscall
jr $ra # Return
##########################################################
#
# BASE 2
#
#########################################################
bin:
sub $sp, $sp, 24 # Push register onto stack
sw $a0, 0($sp)
sw $s0, 4($sp)
sw $s1, 8($sp)
sw $s2, 12($sp)
sw $s3, 16($sp)
sw $s4, 20($sp)
move $s2, $a0 # Move a0 to s2
la $a0, ans1 # Display string before bin answer
li $v0, 4
syscall
li $s0, 32 # 32 digits for base4 word
la $s3, binresult # Bin string set up here
binloop:
rol $s2, $s2, 1 # Start with leftmost digit
and $s1, $s2, 1 # Mask one digit in s2 and place results in s1
binprint:
add $s1, $s1, 48 # Add 48 (30 hex) to get ascii code
sb $s1,($s3) # Store byte in result. s3 -> result
add $s3, $s3, 1 # s3 = s3 + 1
add $s0, $s0, -1 # s0 = s0 - 1
bnez $s0, binloop # If s0 != 0, go to binloop
la $a0, binresult # display result
li $v0, 4
syscall
la $a0, endl
li $v0, 4
syscall
jr $ra # Return
##########################################################
#
# BASE 4
#
#########################################################
base4:
sub $sp, $sp, 24 # Push register onto stack
sw $a0, 0($sp)
sw $s0, 4($sp)
sw $s1, 8($sp)
sw $s2, 12($sp)
sw $s3, 16($sp)
sw $ra, 20($sp)
move $s2, $a0 # Move a0 to s2
la $a0, ans2 # Display string before BASE 4 answer
li $v0, 4
syscall
li $s0, 16 # 16 digits for base4 word
la $s3, base4result # Bin string set up here
base4loop:
rol $s2, $s2, 2 # Start with leftmost digit
and $s1, $s2, 3 # Mask one digit in s2 and place results in s1
fourprint:
add $s1, $s1, 48 # Add 48 (30 hex) to get ascii code
sb $s1,($s3) # Store byte in result. s3 -> result
add $s3, $s3, 1 # s3 = s3 + 1
add $s0, $s0, -1 # s0 = s0 - 1
bnez $s0, base4loop # If s0 != 0, go to binloop
la $a0, base4result # display result
li $v0, 4
syscall
la $a0, endl
li $v0, 4
syscall
jr $ra # Return
##########################################################
#
# BASE 8
#
#########################################################
base8:
sub $sp, $sp, 24 # Push register onto stack
sw $a0, 0($sp)
sw $s0, 4($sp)
sw $s1, 8($sp)
sw $s2, 12($sp)
sw $s3, 16($sp)
sw $ra, 20($sp)
move $s2, $a0 # Move a3 to s2
la $a0, endl
li $v0, 4
syscall
la $a0, ans4 # Display string before bin answer
li $v0, 4
syscall
li $s0, 10 # digits for octal word
la $s3, octresult # Bin string set up here
rol $s2, $s2, 2 # Start with leftmost digit
and $s1, $s2, 0x7 # Mask 7 digits in s2 and place results in s1
base8loop:
rol $s2, $s2, 3 # Start with leftmost digit
and $s1, $s2, 0x7 # Mask 7 digits in s2 and place results in s1
base8print:
add $s1, $s1, 48 # Add 48 (30 hex) to get ascii code
sb $s1,($s3) # Store byte in result. s3 -> result
add $s3, $s3, 1 # s3 = s3 + 1
add $s0, $s0, -1 # s0 = s0 - 1
bnez $s0, base8loop # If s0 != 0, go to binloop
la $a0, octresult # display result
li $v0, 4
syscall
jr $ra # Return
.data
binresult: .space 32
base4result:.space 16
hexresult: .space 8
octresult: .space 10
endl: .asciiz "\n"
prompt: .asciiz "Enter a decimal number: "
ans1: .asciiz "The number in base 2 is "
ans2: .asciiz "The number in base 4 is "
ans3: .asciiz "The number in base 16 is "
ans4: .asciiz "The number in base 8 is "
repeat: .asciiz "Would you like to input another number? "
推荐答案
使用syscall 4打印的字符串必须为ASCIIZ，即带有零终止符的ASCII.因此，您需要在每个字符串的最后一个字符之后存储一个值为零的字节.为了能够存储该额外的字节，您需要为每个字符串保留一个额外的字节(即 binResult 的 .space 33 等).
The strings you print with syscall 4 need to be ASCIIZ, i.e. ASCII with a zero terminator. So you need to store a byte with the value zero after the last character in each string. And to be able to store that extra byte, you need to reserve one additional byte for each string (i.e. .space 33 for binResult, etc).
实际上，仅增加使用 .space 保留的字节数就足够了，如 .space 应该对内存进行零初始化.但是，为确保每个字符串不会造成太大的伤害，每个字符串添加一个额外的 sb .
Actually, just increasing the number of bytes reserved with .space should be enough, as .space should zero-initialize the memory. But adding an extra sb per string just to be sure wouldn't hurt much.
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