Question

我想知道是否有人可以帮我解释如何摆脱mips中字符串中的特定字符。例如，如果字符串是＆＃34; + 104367＆＃34; 我想摆脱+，并简单地说：104367。

Answer 1

 .data
    str:  .asciiz "Hello+World"

 .text  
 main:  
    la $t0, str  # la means load address (so we load the address of str into $t0)  
    li $t1, 0    # $t1 is the counter. set it to 0  

    la $t3, 43   # 43 is ASCII of '+' in DEC
 countChr:  
    lb $t2, 0($t0)  # Load the first byte from address in $t0  
    beqz $t2, end   # if $t2 == 0 then go to label end
    bne $t2, $t3, proceed   # branch if symbol equals 43 (+)
    la $t4, $t0 # save the position
proceed:  
    add $t0, 1      # else increment the address  
    add $t1, $t1, 1 # and increment the counter of course  

    j countChr      # finally loop  


 end:  
    # Do whatever you want here.  
    # Just remember that the length of the string is stored in $t1
    # Here you have the position of '+' saved to t4

如何摆脱mips程序集中字符串中的某些字符

1 个答案: