user: 下面的汇编指令对应的C源代码是什么：

```asm
<split>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  movl   $0x0,(%rcx)
  mov    %rsi,(%rsp)
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     10a0 <split.cold>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rax,%rdx
  movzbl (%rax),%eax
  test   %al,%al
  je     L10
  mov    (%rsp),%r8
  xor    %ebx,%ebx
  xor    %r14d,%r14d
  nopl   (%rax)
L73:
  mov    %ebx,%ebp
  lea    0x1(%rbx),%r15d
  cmp    %al,(%r8)
  je     L80
L65:
  add    $0x1,%rbx
  lea    0x2(%rbp),%r15d
  movzbl (%rdx,%rbx,1),%eax
  test   %al,%al
  jne    L73
  mov    %r15d,%edi
  mov    %rdx,(%rsp)
  sub    %r14d,%edi
  add    $0x1,%edi
  call   <malloc@plt>
  movslq (%r12),%rsi
  mov    %rax,%rcx
  mov    %rax,0x0(%r13,%rsi,8)
  cmp    %r14d,%r15d
  jle    L8
  mov    (%rsp),%rdx
  movslq %r14d,%rsi
  add    %rdx,%rsi
  movzbl (%rsi),%eax
  mov    %al,(%rcx)
  lea    0x1(%r14),%eax
  cmp    %eax,%r15d
  jle    L58
  sub    %r14d,%ebp
  mov    $0x1,%eax
  add    $0x2,%rbp
  cs nopw 0x0(%rax,%rax,1)
L15:
  movslq (%r12),%rdx
  movzbl (%rsi,%rax,1),%ecx
  mov    0x0(%r13,%rdx,8),%rdx
  mov    %cl,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rbp
  jne    L15
L58:
  mov    %r15d,%eax
  mov    $0x1,%ecx
  movslq (%r12),%rdx
  sub    %r14d,%eax
  cmp    %r15d,%r14d
  cmovge %ecx,%eax
  cltq
  add    0x0(%r13,%rdx,8),%rax
L8:
  movb   $0x0,(%rax)
  addl   $0x1,(%r12)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L80:
  mov    %r15d,%edi
  mov    %rdx,0x8(%rsp)
  sub    %r14d,%edi
  mov    %r8,(%rsp)
  call   <malloc@plt>
  movslq (%r12),%rsi
  cmp    %ebx,%r14d
  mov    (%rsp),%r8
  mov    0x8(%rsp),%rdx
  mov    %rax,%rcx
  mov    %rax,0x0(%r13,%rsi,8)
  jge    L29
  movslq %r14d,%rdi
  add    %rdx,%rdi
  movzbl (%rdi),%ecx
  mov    %cl,(%rax)
  lea    0x1(%r14),%eax
  cmp    %ebx,%eax
  jge    L44
  mov    %ebx,%eax
  sub    %r14d,%eax
  lea    -0x2(%rax),%r9d
  mov    $0x1,%eax
  add    $0x2,%r9
  nopl   0x0(%rax,%rax,1)
L70:
  movslq (%r12),%rcx
  movzbl (%rdi,%rax,1),%esi
  mov    0x0(%r13,%rcx,8),%rcx
  mov    %sil,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %r9,%rax
  jne    L70
L44:
  movslq (%r12),%rax
  mov    %ebp,%ecx
  sub    %r14d,%ecx
  movslq %ecx,%rcx
  add    0x0(%r13,%rax,8),%rcx
L29:
  movb   $0x0,(%rcx)
  mov    %r15d,%r14d
  addl   $0x1,(%r12)
  jmp    L65
L10:
  mov    $0x2,%edi
  mov    %rdx,(%rsp)
  xor    %r14d,%r14d
  mov    $0x1,%r15d
  call   <malloc@plt>
  movslq (%r12),%rcx
  mov    (%rsp),%rdx
  mov    %rax,0x0(%r13,%rcx,8)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  jmp    L58
```

assistant: ```c
void split(char *line, char *splitC, char *splitArray[], int *arrayLen) {
  char *line2, c;
  int i, j, start, end;
  start = end = 0;
  *arrayLen = 0;
  line2 = (char *)malloc(strlen(line) + 1);
  if (line2 != NULL)
    line2 = strcpy(line2, line);
  i = 0;
  while ((c = line2[end++]) != '\0') {
    if (c == *splitC) {
      splitArray[*arrayLen] = (char *)malloc((end - start) * sizeof(char));
      i = 0;
      for (j = start; j < end - 1; j++) {
        splitArray[*arrayLen][i++] = line2[j];
      }
      splitArray[*arrayLen][i] = '\0';
      start = end;
      (*arrayLen)++;
    }
  }
  i = 0;
  splitArray[*arrayLen] = (char *)malloc((end - start + 1) * sizeof(char));
  for (j = start; j < end; j++) {
    splitArray[*arrayLen][i++] = line2[j];
  }
  splitArray[*arrayLen][i] = '\0';
  (*arrayLen)++;
}
```