user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<split_string>:
  endbr64
  push   %r14
  mov    %rdi,%rdx
  mov    %esi,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movzbl (%rdi),%eax
  cmp    $0x20,%al
  jne    L207
  nopl   0x0(%rax)
L223:
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  cmp    $0x20,%al
  je     L223
L207:
  cmp    $0x9,%al
  je     L223
  xor    %esi,%esi
  xor    %ebx,%ebx
  mov    $0x8,%edi
  test   %al,%al
  je     L47
  nopl   (%rax)
L119:
  test   $0xdf,%al
  jne    L297
  jmp    L124
  cs nopw 0x0(%rax,%rax,1)
L139:
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  test   $0xdf,%al
  je     L124
L297:
  cmp    $0x9,%al
  jne    L139
L124:
  lea    0x1(%rsi),%ecx
  cmp    $0x20,%al
  jne    L9
  nopw   0x0(%rax,%rax,1)
L77:
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  cmp    $0x20,%al
  je     L77
L9:
  cmp    $0x9,%al
  je     L77
  test   %al,%al
  je     L259
  mov    %ecx,%esi
  jmp    L119
  nopl   0x0(%rax,%rax,1)
L259:
  movslq %ecx,%rcx
  lea    0x10(,%rsi,8),%edi
  lea    0x0(,%rcx,8),%rbx
L47:
  call   <smalloc@plt>
  movq   $0x0,(%rax,%rbx,1)
  mov    %rax,%r13
  movzbl 0x0(%rbp),%eax
  cmp    $0x20,%al
  jne    L48
  nopw   0x0(%rax,%rax,1)
L94:
  movzbl 0x1(%rbp),%eax
  add    $0x1,%rbp
  cmp    $0x20,%al
  je     L94
L48:
  cmp    $0x9,%al
  je     L94
  test   %al,%al
  je     L86
  mov    %r13,%r12
  test   $0xdf,%al
  je     L310
  nopl   0x0(%rax,%rax,1)
L106:
  mov    %rbp,%rbx
  cmp    $0x9,%al
  jne    L275
  jmp    L310
  nopl   0x0(%rax)
L212:
  cmp    $0x9,%al
  je     L298
L275:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  test   $0xdf,%al
  jne    L212
L298:
  test   %r14d,%r14d
  jne    L229
L165:
  mov    %rbp,(%r12)
  cmpb   $0x0,(%rbx)
  jne    L289
L86:
  pop    %rbx
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L289:
  movb   $0x0,(%rbx)
L294:
  movzbl 0x1(%rbx),%eax
  lea    0x1(%rbx),%rbp
  cmp    $0x9,%al
  je     L83
  cmp    $0x20,%al
  jne    L210
  nopl   0x0(%rax,%rax,1)
L83:
  movzbl 0x1(%rbp),%eax
  add    $0x1,%rbp
  cmp    $0x20,%al
  je     L83
  cmp    $0x9,%al
  je     L83
L210:
  add    $0x8,%r12
  test   %al,%al
  je     L86
  test   $0xdf,%al
  jne    L106
L310:
  mov    %rbp,%rbx
  test   %r14d,%r14d
  je     L165
  nopl   0x0(%rax)
L229:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  sub    %ebp,%esi
  call   <string_ndup@plt>
  mov    %rax,(%r12)
  cmpb   $0x0,(%rbx)
  jne    L294
  pop    %rbx
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char **split_string(char *str, int copy) {
  char **bits;
  int nbits;
  char *ch, *start;
  for (ch = str; (*ch == ' ') || (*ch == '\t'); ch++)
    ;
  for (nbits = 0; *ch != '\0';) {
    for (; (*ch != '\0') && (*ch != ' ') && (*ch != '\t'); ch++)
      ;
    nbits++;
    for (; (*ch == ' ') || (*ch == '\t'); ch++)
      ;
  }
  bits = (char **)smalloc((nbits + 1) * sizeof(char *));
  bits[nbits] = NULL;
  for (ch = str; (*ch == ' ') || (*ch == '\t'); ch++)
    ;
  for (nbits = 0; *ch != '\0';) {
    start = ch;
    for (; (*ch != '\0') && (*ch != ' ') && (*ch != '\t'); ch++)
      ;
    if (copy) {
      bits[nbits] = string_ndup(start, (int)(ch - start));
      if (*ch != '\0') {
        ch++;
      }
    } else {
      bits[nbits] = start;
      if (*ch != '\0') {
        *ch = '\0';
        ch++;
      }
    }
    nbits++;
    for (; (*ch == ' ') || (*ch == '\t'); ch++)
      ;
  }
  return bits;
}
```