user: 请从给定的汇编指令推导出C源代码：

```asm
<myAtoi>:
  endbr64
  test   %rdi,%rdi
  je     L191
  mov    %rdi,%rcx
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L90
  push   %rbp
  push   %rbx
  mov    D8(%rip),%rax
  mov    (%rax),%r10d
  movslq %r10d,%r8
  imul   $0x66666667,%r8,%r8
  sar    $0x22,%r8
  mov    %r10d,%eax
  sar    $0x1f,%eax
  sub    %eax,%r8d
  mov    D2(%rip),%rax
  mov    (%rax),%r9d
  movslq %r9d,%rdi
  imul   $0x66666667,%rdi,%rdi
  sar    $0x22,%rdi
  mov    %r9d,%eax
  sar    $0x1f,%eax
  sub    %eax,%edi
  mov    $0x0,%r11d
  mov    $0x0,%eax
  mov    $0x0,%ebx
  mov    $0x1,%esi
  jmp    L62
L43:
  add    $0x1,%rcx
  jmp    L246
L219:
  lea    0x1(%rcx),%rbx
  movzbl 0x1(%rcx),%edx
  test   %dl,%dl
  je     L2
  lea    -0x30(%rdx),%ecx
  cmp    $0x9,%cl
  ja     L34
  mov    %rbx,%rcx
L187:
  cmp    %eax,%edi
  jg     L71
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  movsbl %dl,%edx
  sub    $0x30,%edx
  lea    (%r9,%rdx,1),%r11d
  cmp    %eax,%r11d
  jg     L221
  sub    %edx,%eax
  add    $0x1,%rcx
  mov    %esi,%r11d
  mov    %esi,%ebx
  jmp    L246
L36:
  movzbl (%rcx),%edx
  lea    -0x30(%rdx),%r11d
  cmp    $0x9,%r11b
  ja     L34
  test   %ebx,%ebx
  jne    L187
  cmp    %eax,%r8d
  jl     L195
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  movsbl %dl,%edx
  sub    $0x30,%edx
  mov    %r10d,%r11d
  sub    %edx,%r11d
  cmp    %eax,%r11d
  jl     L186
  add    %edx,%eax
  add    $0x1,%rcx
  mov    %esi,%r11d
L246:
  movzbl (%rcx),%edx
  test   %dl,%dl
  je     L34
L62:
  test   %r11d,%r11d
  jne    L36
  mov    %edx,%ebp
  and    $0xffffffef,%ebp
  cmp    $0x20,%bpl
  je     L43
  cmp    $0x2d,%dl
  je     L219
  cmp    $0x2b,%dl
  jne    L36
  lea    0x1(%rcx),%rdx
  cmpb   $0x0,0x1(%rcx)
  je     L179
  mov    %rdx,%rcx
  jmp    L36
L191:
  mov    $0x0,%eax
  ret
L90:
  mov    $0x0,%eax
  ret
L2:
  mov    %r11d,%eax
  jmp    L34
L179:
  mov    %r11d,%eax
  jmp    L34
L195:
  mov    %r10d,%eax
L34:
  pop    %rbx
  pop    %rbp
  ret
L186:
  mov    %r10d,%eax
  jmp    L34
L71:
  mov    %r9d,%eax
  jmp    L34
L221:
  mov    %r9d,%eax
  jmp    L34
```

assistant: ```c
int myAtoi(char *str) {
  if (str == NULL)
    return 0;
  int negtive = 0;
  int result = 0;
  int tmp = 0;
  int start = 0;
  char *p = str;
  while (*p) {
    if (start == 0) {
      if (*p == ' ' || *p == '0') {
        p++;
        continue;
      } else {
        start = 1;
        if (*p == '-') {
          negtive = 1;
          p++;
          if (*p == '\0') {
            return 0;
          }
        } else if (*p == '+') {
          p++;
          if (*p == '\0') {
            return 0;
          }
        }
      }
    }
    if (*p >= '0' && *p <= '9') {
      if (!negtive) {
        if (INT_MAX / 10 >= result) {
          result *= 10;
          tmp = *p - '0';
          if (INT_MAX - tmp >= result) {
            result += tmp;
            p++;
          } else {
            return INT_MAX;
          }
        } else {
          return INT_MAX;
        }
      } else {
        if (INT_MIN / 10 <= result) {
          result *= 10;
          tmp = *p - '0';
          if (INT_MIN + tmp <= result) {
            result -= tmp;
            p++;
          } else {
            return INT_MIN;
          }
        } else {
          return INT_MIN;
        }
      }
    } else {
      break;
    }
  }
  return result;
}
```