user: What would the C source look like for the assembly code provided here?

```asm
<__itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbx
  mov    %edi,%ebx
  sub    $0x68,%rsp
  mov    %ecx,-0x7c(%rbp)
  mov    %rsi,-0x88(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  mov    %ecx,%eax
  mov    %rsp,%rcx
  add    $0x1,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L139
L97:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L97
L139:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L53
L25:
  movdqa D10(%rip),%xmm0
  lea    -0x2(%r12),%eax
  mov    %rsp,-0x78(%rbp)
  movaps %xmm0,-0x70(%rbp)
  cmp    $0xe,%eax
  ja     L140
  cmpq   $0x0,-0x88(%rbp)
  je     L140
  xor    %r15d,%r15d
  lea    -0x60(%rbp),%r13
  mov    $0x1f,%r14d
  test   %ebx,%ebx
  je     L55
  nopl   0x0(%rax)
L75:
  mov    %r14d,%esi
  mov    %r12d,%edi
  sub    %r15d,%esi
  call   <pow@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  cltq
  mov    %edx,%ebx
  movzbl -0x70(%rbp,%rax,1),%eax
  mov    %al,0x0(%r13,%r15,1)
  add    $0x1,%r15
  cmp    $0x20,%r15
  jne    L75
  mov    -0x7c(%rbp),%ecx
  mov    $0x20,%eax
  sub    %ecx,%eax
  cmp    $0x1f,%eax
  jg     L132
  cltq
  mov    -0x78(%rbp),%rdi
  mov    %ecx,%edx
  lea    0x0(%r13,%rax,1),%rsi
  call   <memcpy@plt>
L132:
  movslq -0x7c(%rbp),%rax
  mov    -0x78(%rbp),%rdx
  movb   $0x0,(%rdx,%rax,1)
  movzbl (%rdx),%eax
  test   %al,%al
  jne    L14
  jmp    L48
  nopl   (%rax)
L145:
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L48
L14:
  cmp    $0x30,%al
  je     L145
  mov    %rdx,-0x78(%rbp)
L48:
  mov    -0x78(%rbp),%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rbx,%rsi
  mov    -0x88(%rbp),%rbx
  lea    0x1(%rax),%rdx
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rax
L58:
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L65
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L140:
  xor    %eax,%eax
  jmp    L58
L53:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L25
L55:
  mov    -0x88(%rbp),%rbx
  lea    D5(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  jmp    L58
L65:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D10
type_infered: byte[16]
name: unpack_data
arguments: label: D5
type_infered: string

tool: {"D10": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x61,0x62,0x63,0x64,0x65,0x66]}

tool: D5: 0

assistant: ```c
char *__itoa(int val, char *result, int base, int nbrNum) {
  int j = 0;
  int i = 0;
  int k = 0;
  int l = 0;
  int temp;
  char *p, *p2;
  char res[32];
  char res2[nbrNum + 1];
  char cList[] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'};
  if (result == NULL || base > 16 || base < 2)
    return NULL;
  if (0 == val) {
    strcpy(result, "0");
    return result;
  }
  j = val;
  i = 31;
  while (i >= 0) {
    l = pow(base, i);
    temp = (int)j / l;
    res[k++] = cList[temp];
    j -= temp * l;
    i--;
  }
  i = 0;
  j = 31 - (nbrNum - 1);
  for (k = j; k < 32; k++) {
    res2[i++] = res[k];
  }
  res2[nbrNum] = '\0';
  p = p2 = res2;
  while (*p) {
    if (*p != '0') {
      p2 = p;
      break;
    }
    p++;
  }
  memcpy(result, p2, strlen(p2) + 1);
  return result;
}
```