user: 请从给定的汇编指令推导出C源代码:

```asm
<itos>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x24(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L84
L70:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x1c(%rbp)
  addl   $0x1,-0x20(%rbp)
L84:
  cmpl   $0x0,-0x1c(%rbp)
  jg     L70
  cmpl   $0x0,-0x20(%rbp)
  jg     L75
  movl   $0x1,-0x20(%rbp)
L75:
  cmpl   $0x0,-0x24(%rbp)
  jns    L13
  mov    $0x2,%edx
  jmp    L95
L13:
  mov    $0x1,%edx
L95:
  mov    -0x20(%rbp),%eax
  add    %eax,%edx
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x0,-0x24(%rbp)
  jns    L71
  mov    $0x2,%edx
  jmp    L58
L71:
  mov    $0x1,%edx
L58:
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jns    L52
  mov    -0x10(%rbp),%rax
  movb   $0x2d,(%rax)
  addq   $0x1,-0x10(%rbp)
L52:
  mov    -0x20(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  jmp    L14
L3:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%edi
  mov    -0x14(%rbp),%eax
  cltd
  idiv   %edi
  mov    %eax,%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  movslq %edx,%rax
  lea    D24(%rip),%rdx
  movzbl (%rax,%rdx,1),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x10(%rbp)
  subl   $0x1,-0x18(%rbp)
L14:
  cmpl   $0x0,-0x18(%rbp)
  jns    L3
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
data_type: string
name: disassemble_data
arguments: data_tag: D17
data_type: string
name: disassemble_data
arguments: data_tag: D24
data_type: string
name: disassemble_data
arguments: data_tag: D19
data_type: string

tool: D3: --- invoking itos(%d) ---


tool: D17:  - size=%d, minus= ==> malloc(%d)


tool: D24: 0123456789

tool: D19: --- finished itos(%d): %s ---


assistant: ```c
char *itos(int i) {
  printf("--- invoking itos(%d) ---\n", i);
  uint u = (i < 0 ? i * -1 : i);
  uint size = 0;
  uint j = u;
  while (j > 0) {
    j /= 10;
    size++;
  }
  if (size < 1)
    size = 1;
  printf(" - size=%d, minus= ==> malloc(%d)\n", size, (size + (i < 0 ? 2 : 1)));
  char *s = malloc(size + (i < 0 ? 2 : 1));
  char *p = s;
  if (i < 0) {
    *p = '-';
    p++;
  }
  for (int k = size - 1; k >= 0; k--) {
    *p = "0123456789"[(u / (int)pow(10, k)) % 10];
    p++;
  }
  *p = 0;
  printf("--- finished itos(%d): %s ---\n", i, s);
  return s;
}
```