Assembly MIPS汇编代码，用于计算字符串中的位数

我正在编写一个MIPS汇编代码来计算字符串中的位数。例如，如果用户字符串为：“qapww9…$$$$64” 输出为3

我使用ascii码设置整数的边界（48代表0，57代表9）。我想检查每个字符是否大于或等于0，然后小于或等于9，如果是，则在计数器中添加一个，然后移动到下一个字符。键入字符串输入后，我的代码失败

编辑 ：我收到的反馈是，我没有正确地递增字符串的每个字符。我目前正在做的是

添加$t0，$t0，1

# MIPS assembly language program to count the number of decimal digits in an ascii string of characters.
# $a0 = $t0 = user input
# $t1 = counter
# $t2 = temporary register to hold each byte value
.data
    length: .space 100
    p: .asciiz "\nEnter a string: "
    p2: .asciiz "\nNumber of integers: "
.text

#Prompt User for String
   la $a0, p                   #print prompt
   li $v0, 4                #load string stored in v0
   syscall                  #display prompt string

#Get user input
   li $v0, 8                #load string stored in v0
   la $a0, length              #set string length to 100
   addi $a1, $0, 100           #add length 100 to regist $a1
   syscall

   move $t0, $a0                #user input is now in register $t0
   addi $t1, $0, 0              #initialize counter

#loop to check if greater or equal to 0
lowerBound:
   lb $t2, 0($t0)           #load first character of user input into $t2
   bge $t2, 48, upperBound     #branch to upperBound checker if char is greater than or equal 0
   addi $t0, $t0, 1         #increment to next character in string
   beqz $t0, end            #if character = 0 (end of string), end
   j lowerBound

#loop to check if less than or equal to 9
upperBound:
    ble  $t2, 57, count_digits  #if in the range 0-9, just to count digits and add 1 to counter
    j lowerBound            #loop through again

count_digits:
    addi $t1, $t1, 1           #add 1 to counter
    j lowerBound

end:
    li $v0, 4          #load string print service
    la $a0, p2         #load address of prompt 2 inro $a0
    syscall            #print prompt
    la $a0, ($t1)          #load address of count into $a0
    li $v0, 1          #specify print integer service
    syscall            #print count

---

当字符ASCII大于57时，您陷入了无休止的循环。因为在本例中，由于指向字符串（$t0）的指针在本例中不会增加，因此您将无休止地跳转到

上限

，在这里您将跳回

下限

。只需将增量部分移到

bge

上方，它就会一直执行

此外，由于未知原因，您尝试打印计数地址，而不是您应该打印的值

修改后的代码

# MIPS assembly language program to count the number of decimal digits in an ascii string of characters.
# $a0 = $t0 = user input
# $t1 = counter
# $t2 = temporary register to hold each byte value
.data
    length: .space 100
    p: .asciiz "\nEnter a string: "
    p2: .asciiz "\nNumber of integers: "
.text

#Prompt User for String
   la $a0, p                   #print prompt
   li $v0, 4                #load string stored in v0
   syscall                  #display prompt string

#Get user input
   li $v0, 8                #load string stored in v0
   la $a0, length              #set string length to 100
   addi $a1, $0, 100           #add length 100 to regist $a1
   syscall

   move $t0, $a0                #user input is now in register $t0
   addi $t1, $0, 0              #initialize counter

#loop to check if greater or equal to 0
lowerBound:
   lb $t2, 0($t0)           #load first character of user input into $t2
   addi $t0, $t0, 1         #increment to next character in string
   bge $t2, 48, upperBound     #branch to upperBound checker if char is greater than or equal 0
   beqz $t2, end            #if character = 0 (end of string), end
   j lowerBound

#loop to check if less than or equal to 9
upperBound:
    ble  $t2, 57, count_digits  #if in the range 0-9, just to count digits and add 1 to counter
    j lowerBound            #loop through again

count_digits:
    addi $t1, $t1, 1           #add 1 to counter
    j lowerBound

end:
    li $v0, 4          #load string print service
    la $a0, p2         #load address of prompt 2 inro $a0
    syscall            #print prompt
    move $a0, $t1          #load count into $a0
    li $v0, 1          #specify print integer service
    syscall            #print count

---

您正在反复加载同一个字符。您的计数器不是

$t2

，您打算如何准确地到达

末尾？哪里出错了？哪里没有递增字符串中要检查的每个字符。我计划用这句话结束：beqz$t2，结束。。。我相信这意味着如果字符$t2=0，那么这意味着字符串是空的，因此我完成了检查。
lb$t2，0（$t0）
正在从
$t0
加载，而
$t0
从未更改。此外，如果字符小于48，您只需在
下半框中循环，就永远不会退出。
addi$t2，$t2，1
这是否会移动到下一个字符？然后它再次循环通过
下一个
来检查下一个？
$t2
是您加载的字符。增加它是没有意义的。您希望递增
$t0
，以便可以使用它将下一个字符加载到
$t2
中。