以mips连接两个字符串的字母

Question

您好一直试图连接到字符串输入是两个字符串：“john”和“la”，它应该打印“lala”，而不是打印“llll”，我无法修复它。

.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

其实，如果你想在一个字符逐个字符的基础两个字符串合并，正确的输出是jloahn

您的代码甚至不会安装。而且，它只会使用两个字符串中的一个。我提供了两个版本。其中一个注释了错误。另一个是完全正确的版本[请原谅无偿风格的清理]。

的注释版本：

.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 

的清理和纠正版：

.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