user: 这段汇编代码用C语言表示是什么样的?

```asm
<edflib_get_long_time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  addq   $0x1,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L9
L79:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L36
  movl   $0x1,-0x1c(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  jmp    L105
L36:
  addl   $0x1,-0x20(%rbp)
L9:
  mov    -0x20(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L79
L105:
  cmpl   $0x0,-0x1c(%rbp)
  je     L125
  movq   $0x989680,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x20(%rbp)
  jmp    L94
L118:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  imul   -0x8(%rbp),%rax
  add    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,-0x8(%rbp)
  subl   $0x1,-0x20(%rbp)
L94:
  cmpl   $0x0,-0x20(%rbp)
  jns    L118
  movq   $0xf4240,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x20(%rbp)
  jmp    L68
L29:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  imul   -0x8(%rbp),%rax
  add    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  addl   $0x1,-0x20(%rbp)
L68:
  mov    -0x20(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L29
  jmp    L108
L125:
  movq   $0x989680,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x20(%rbp)
  jmp    L25
L32:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  imul   -0x8(%rbp),%rax
  add    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,-0x8(%rbp)
  subl   $0x1,-0x20(%rbp)
L25:
  cmpl   $0x0,-0x20(%rbp)
  jns    L32
L108:
  mov    -0x28(%rbp),%rax
  sub    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L110
  negq   -0x10(%rbp)
L110:
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
long long edflib_get_long_time(char *str) {
  int i, len, hasdot = 0, dotposition = 0;
  long long value = 0, radix;
  str = str + 1;
  len = strlen(str);
  for (i = 0; i < len; i++) {
    if (str[i] == '.') {
      hasdot = 1;
      dotposition = i;
      break;
    }
  }
  if (hasdot) {
    radix = (10000000LL);
    for (i = dotposition - 1; i >= 0; i--) {
      value += ((long long)(str[i] - 48)) * radix;
      radix *= 10;
    }
    radix = (10000000LL) / 10;
    for (i = dotposition + 1; i < len; i++) {
      value += ((long long)(str[i] - 48)) * radix;
      radix /= 10;
    }
  } else {
    radix = (10000000LL);
    for (i = len - 1; i >= 0; i--) {
      value += ((long long)(str[i] - 48)) * radix;
      radix *= 10;
    }
  }
  if (str[-1] == '-')
    value = -value;
  return (value);
}
```