如何将字符串复制到64位ARM组件中动态分配的新字符串中？

Question

我将如何在下面的程序中动态分配数据，而不是静态分配。至少从我听到的内容中，您需要遵循3个步骤...

1. malloc静态字符串
2. 获取字符串的长度，然后将其传递给Malloc
3. 将字符串复制到新的malloc'd字符串中。 （？）

如何完成步骤3？这是我到目前为止的东西...

.data 

str1:   .asciz "The Cat in the Hat\n"

data:   .quad 0

.global _start

.text

_start:

    // Static Version

    LDR X0,=str1;   // Points to the string "Cat in the Hat\n" 

    LDR X1,=data;   // Points to "Data"
    STR X0,[X1].    // Stores the string into "Data"


    // Dynamic Version: What I have so far...
    
    LDR X0,=str1;   // Points to the string "Cat in the Hat\n" 

    BL  strlength   // Returns the length of the string in X0. Modifies registers 
                    // X0, X1, X2, and X7 
    ADD X0,X0,#1    // +1 to the length to account for the null at the end 
    BL  malloc      // Passes X0 into malloc as the amount of bytes to be dynamically 
                    // allocated. Returns the Address of the block of memory allocated 
                    // in X0

    LDR X1,=str1    // Holds the string 

loop:

    LDRB    W5,[X1],#1  // Grabs the first byte from the string
    STRB    W5,[X0],#1  // Stores that byte into X0

    CMP     W5, #0      // Checks for Null
    B.EQ    endLoop     // Branches if Null
    
    B       loop        // Loops
    

    endLoop:       
          
                                        


    LDR X1,=data;   // Points to "Data"
    STR X0,[X1].    // Stores the string into "Data"


end:
    MOV X0, #0
    MOV X8, #93
    SVC 0