user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<split>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdx,%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,%r14
  movzbl (%rax),%eax
  test   %al,%al
  je     L64
  mov    (%rsp),%r8
  xor    %ebx,%ebx
  xor    %r15d,%r15d
  nopl   (%rax)
L45:
  mov    %ebx,%ebp
  lea    0x1(%rbx),%r9d
  cmp    %al,(%r8)
  je     L12
L107:
  add    $0x1,%rbx
  add    $0x2,%ebp
  movzbl (%r14,%rbx,1),%eax
  test   %al,%al
  jne    L45
  mov    %ebp,%edi
  sub    %r15d,%edi
  add    $0x1,%edi
  call   <malloc@plt>
  movslq 0x0(%r13),%rdx
  mov    %rax,(%r12,%rdx,8)
  cmp    %ebp,%r15d
  jge    L16
L39:
  lea    0x1(%r15),%edx
  movslq %r15d,%rcx
  movslq %edx,%rdx
  jmp    L65
  nopl   (%rax)
L108:
  movslq 0x0(%r13),%rax
  add    $0x1,%rdx
  mov    (%r12,%rax,8),%rax
L65:
  movzbl -0x1(%r14,%rdx,1),%edi
  mov    %rdx,%rsi
  sub    %rcx,%rsi
  mov    %dil,-0x1(%rax,%rsi,1)
  cmp    %edx,%ebp
  jg     L108
  mov    %ebp,%eax
  mov    $0x1,%ecx
  movslq 0x0(%r13),%rdx
  sub    %r15d,%eax
  cmp    %r15d,%ebp
  cmovle %ecx,%eax
  cltq
  add    (%r12,%rdx,8),%rax
L16:
  movb   $0x0,(%rax)
  addl   $0x1,0x0(%r13)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  mov    %r9d,%edi
  mov    %r8,0x8(%rsp)
  sub    %r15d,%edi
  mov    %r9d,(%rsp)
  call   <malloc@plt>
  cmp    %ebx,%r15d
  mov    (%rsp),%r9d
  mov    0x8(%rsp),%r8
  mov    %rax,%rdx
  movslq 0x0(%r13),%rax
  mov    %rdx,(%r12,%rax,8)
  jge    L82
  mov    %ebx,%r10d
  movslq %r15d,%rcx
  xor    %eax,%eax
  sub    %r15d,%r10d
  add    %r14,%rcx
  lea    -0x1(%r10),%edi
  jmp    L77
  nopl   0x0(%rax)
L1:
  movslq 0x0(%r13),%rdx
  add    $0x1,%rax
  mov    (%r12,%rdx,8),%rdx
L77:
  movzbl (%rcx,%rax,1),%esi
  mov    %sil,(%rdx,%rax,1)
  cmp    %rax,%rdi
  jne    L1
  movslq 0x0(%r13),%rax
  movslq %r10d,%r10
  mov    (%r12,%rax,8),%rdx
  add    %r10,%rdx
L82:
  movb   $0x0,(%rdx)
  mov    %r9d,%r15d
  addl   $0x1,0x0(%r13)
  jmp    L107
L64:
  mov    $0x2,%edi
  mov    $0x1,%ebp
  xor    %r15d,%r15d
  call   <malloc@plt>
  movslq 0x0(%r13),%rdx
  mov    %rax,(%r12,%rdx,8)
  jmp    L39
```

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)++;
}
```