Question

你好我一直试图连接字符串输入是两个字符串：＆＃34; john＆＃34;和＆＃34; la＆＃34;它应该打印＆＃34; lala＆＃34;相反，它打印＆＃34; llll＆＃34;我无法解决它。

.data  
texto:  .asciiz "john"
cifra:  .asciiz "la"
.text

la $a0,texto  #pass address of str1
la $a1,cifra  #pass address of str2

jal conc
conc:
add $t1,$zero,$a0
add $t2,$zero,$a1


loop:
lb $t3($t1)  
lb $t4($t2) 
lbu $t4,0($a1)  
sb $t4, 0($a1)

beq $t3, $zero, end
nop
move $a0,$t4
li   $v0, 11
syscall
addi $t1, $t1, 1
addi $t2,$t2, 1
j loop

end:

Answer 1

实际上，如果你想在逐个字符的基础上合并两个字符串，那么正确的输出是jloahn

您的代码甚至无法汇编。而且，它只会使用两个字符串中的一个。我提供了两个版本。一个是用错误注释的。另一个是完全修正的版本[请原谅无偿的风格清理]。

带注释的版本：

    .data
texto:      .asciiz     "john"
cifra:      .asciiz     "la"
    .text

    la      $a0,texto               # pass address of str1
    la      $a1,cifra               # pass address of str2

    jal     conc

    # BUG: no exit program
    li      $v0,10
    syscall

conc:
    add     $t1,$zero,$a0
    add     $t2,$zero,$a1

loop:
    # BUG: these won't even assemble
    ###lb       $t3($t1)
    ###lb       $t4($t2)

    lb      $t3,0($t1)
    lb      $t4,0($t2)

    # BUG: these fetch from $a1 which is never incremented -- in other words,
    # t4 will always get the same first char (i.e. "l")
    ###lbu      $t4,0($a1)
    ###sb       $t4,0($a1)

    beq     $t3,$zero,end
    nop

    # output next char
    # BUG: this only outputs chars from one string
    move    $a0,$t4
    li      $v0,11
    syscall

    addi    $t1,$t1,1
    addi    $t2,$t2,1
    j       loop

end:
    jr      $ra

已清理并更正的版本：

    .data
texto:      .asciiz     "john"
cifra:      .asciiz     "la"
out:        .space      80

    .text

    la      $a0,out                 # get address of output

    la      $a1,texto               # pass address of str1
    la      $a2,cifra               # pass address of str2
    jal     conc

    # output the concatenated string
    la      $a0,out
    li      $v0,4
    syscall

    # BUG: no exit program
    li      $v0,10
    syscall

# conc -- concatenate strings char-by-char
#
# arguments:
#   a0 -- output pointer
#   a1 -- pointer to string to concatenate
#   a2 -- pointer to string to concatenate
conc:
    lb      $t1,0($a1)              # get string char
    beqz    $t1,conc_no1st          # EOS? if yes, skip
    addi    $a1,$a1,1               # advance 1st source pointer
    sb      $t1,0($a0)              # add to output
    addi    $a0,$a0,1               # advance output pointer

conc_no1st:
    lb      $t2,0($a2)              # add to output
    beqz    $t2,conc_no2nd          # EOS? if yes, skip
    addi    $a2,$a2,1               # advance 2nd source pointer
    sb      $t2,0($a0)              # add to output
    addi    $a0,$a0,1               # advance output pointer

conc_no2nd:
    or      $t1,$t1,$t2             # both chars EOS?
    bnez    $t1,conc                # no, loop

    sb      $zero,0($a0)            # store final EOS
    jr      $ra

用mips

1 个答案: