為什么我的 MIPS 基本轉換器在當前值之后打印出前一個循環中的值？
[英]Why is my MIPS base converter printing out the values from a previous loop after the current values?
問題描述
我對 MIPS 很陌生，這讓我完全困惑。 我制作了一個轉換基數的程序，它第一次運行良好，但是當它循環時，它會顯示來自循環先前迭代的其他寄存器的值。 輸出如下。 我已經嘗試了我能想到的一切，但我沒有想法......
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
您想輸入另一個號碼嗎？ 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
您想輸入另一個號碼嗎？
.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? "
2 個解決方案
解決方案1
2 2016-05-11 05:22:35
您使用 syscall 4 打印的字符串必須是 ASCIIZ，即帶有零終止符的 ASCII。 因此，您需要在每個字符串的最后一個字符后存儲一個值為 0 的字節。 為了能夠存儲額外的字節，您需要為每個字符串保留一個額外的字節（即binResult等的.space 33 ）。
其實，只是增加了與保留的字節數.space應該是不夠的，因為.space應該零初始化的內存。 但是為每個字符串添加一個額外的sb只是為了確保不會受到太大傷害。
解決方案2
1 2016-05-11 06:17:33
我在理解所有帶有一些沖突注釋和寄存器用法的代碼時遇到了一些困難（例如，輸入 27 並且十六進制返回 0）。 當我運行它，通過它加強，我想我看到了rol垃圾的東西。 不知道，因為我只運行了一次或兩次。
所以，我使用不同的方法重新編碼。 所有四個鹼基的代碼現在是通用的。 我做了十六進制和八進制。 我把基地 4 和基地 2 留給你去做。
無論如何，這是代碼[請原諒無償的風格清理]：
.text
.globl main
main:
la $a0,prompt # Prompt for a base10 integer
li $v0,4
syscall
# get the value
li $v0,5
syscall
move $s7,$v0
###jal bin
###jal base4
jal hex
jal oct
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 main
eop:
li $v0,10 # End Of Program
syscall
# BASE 16
hex:
la $a0,hexmsg # Display string before hex answer
li $a1,0x0F # mask for hex digit
li $a2,28 # initial right shift amount
li $a3,4 # right shift decrement
j numdump
# BASE 8
oct:
la $a0,octmsg # Display string before hex answer
li $a1,0x07 # mask for octal digit
li $a2,30 # right shift amount
li $a3,3 # right shift decrement
j numdump
# numdump -- dump out a number in an alternate base
#
# arguments:
# a0 -- pointer to string for prefix
# a1 -- mask for digit
# a2 -- initial right shift amount
# a3 -- amount to decrement shift by
# s7 -- number value
numdump:
li $v0,4
syscall
la $t3,result # output string set up here
numloop:
srlv $t0,$s7,$a2 # slide the digit right
and $t0,$t0,$a1 # mask the digit
lb $t0,digits($t0) # get the ascii value
sb $t0,0($t3) # store into result buffer
addi $t3,$t3,1 # advance result pointer
sub $a2,$a2,$a3 # reduce shift amount -- more to do?
bgez $a2,numloop # yes, loop
sb $zero,0($t3) # store end of string
la $a0,result # display result
li $v0,4
syscall
la $a0,endl
syscall
jr $ra # Return
.data
result: .space 40
digits: .asciiz "0123456789ABCDEF"
endl: .asciiz "\n"
prompt: .asciiz "Enter a decimal number: "
b2msg: .asciiz "The number in base 2 is "
b4msg: .asciiz "The number in base 4 is "
hexmsg: .asciiz "The number in base 16 is "
octmsg: .asciiz "The number in base 8 is "
repeat: .asciiz "Would you like to input another number? "
這是一個稍微更緊湊的版本，它需要對 common 函數少一個參數：
.text
.globl main
main:
la $a0,prompt # Prompt for a base10 integer
li $v0,4
syscall
# get the value
li $v0,5
syscall
move $s7,$v0
###jal bin
###jal base4
jal hex
jal oct
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 main
eop:
li $v0,10 # End Of Program
syscall
# BASE 16
hex:
la $a0,hexmsg # Display string before hex answer
li $a1,4 # number of bits in a digit
li $a2,28 # initial right shift amount
j numdump
# BASE 8
oct:
la $a0,octmsg # Display string before hex answer
li $a1,3 # number of bits in a digit
li $a2,30 # right shift amount
j numdump
# numdump -- dump out a number in an alternate base
#
# arguments:
# a0 -- pointer to string for prefix
# a1 -- number of bits in a digit
# a2 -- initial right shift amount
# s7 -- number value
#
# registers:
# a3 -- digit mask
numdump:
li $v0,4
syscall
la $t3,result # output string set up here
# create digit mask from number of bits in a digit
li $a3,1 # mask = 1
sllv $a3,$a3,$a1 # mask <<= digit width (for hex, 0x10)
subiu $a3,$a3,1 # bump down for mask (for hex, 0x0F)
numloop:
srlv $t0,$s7,$a2 # slide the digit right
and $t0,$t0,$a3 # mask the digit
lb $t0,digits($t0) # get the ascii value
sb $t0,0($t3) # store into result buffer
addi $t3,$t3,1 # advance result pointer
sub $a2,$a2,$a1 # reduce shift amount -- more to do?
bgez $a2,numloop # yes, loop
sb $zero,0($t3) # store end of string
la $a0,result # display result
li $v0,4
syscall
la $a0,endl
syscall
jr $ra # Return
.data
result: .space 40
digits: .asciiz "0123456789ABCDEF"
endl: .asciiz "\n"
prompt: .asciiz "Enter a decimal number: "
b2msg: .asciiz "The number in base 2 is "
b4msg: .asciiz "The number in base 4 is "
hexmsg: .asciiz "The number in base 16 is "
octmsg: .asciiz "The number in base 8 is "
repeat: .asciiz "Would you like to input another number? "
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