user: 这段汇编代码用C语言表示是什么样的？

```asm
<Logarithm>:
  endbr64
  push   %r12
  lea    D13(%rip),%rdi
  push   %rbp
  lea    D27(%rip),%rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r12
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    $0x1,%eax
  je     L65
  cmp    $0x2,%eax
  je     L38
  cmp    $0x3,%eax
  je     L36
L47:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L59
  add    $0x28,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L65:
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%edi
  call   <log10@plt>
  mov    0xc(%rsp),%esi
  mov    $0x1,%eax
  lea    D61(%rip),%rdi
  call   <printf@plt>
  jmp    L47
  nopl   0x0(%rax,%rax,1)
L38:
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%ebp
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebp,%xmm0
  call   <log@plt>
  mov    %ebp,%esi
  mov    $0x1,%eax
  lea    D31(%rip),%rdi
  call   <printf@plt>
  jmp    L47
  nopl   0x0(%rax)
L36:
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D52(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%r12d
  pxor   %xmm0,%xmm0
  cvtsi2sd %r12d,%xmm0
  call   <log@plt>
  movq   %xmm0,%rbp
  movsd  0x10(%rsp),%xmm0
  call   <log@plt>
  movapd %xmm0,%xmm1
  movq   %rbp,%xmm0
  divsd  %xmm1,%xmm0
  call   <log@plt>
  mov    %r12d,%esi
  mov    $0x1,%eax
  lea    D31(%rip),%rdi
  call   <printf@plt>
  jmp    L47
L59:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D13
type_infered: string
name: read_data
arguments: data_tag: D27
type_infered: string
name: read_data
arguments: data_tag: D3
type_infered: string
name: read_data
arguments: data_tag: D61
type_infered: string
name: read_data
arguments: data_tag: D31
type_infered: string
name: read_data
arguments: data_tag: D38
type_infered: string
name: read_data
arguments: data_tag: D52
type_infered: string

tool: D13: 1-Compute log base 10 
2- calculate natural logarithm 
3-Custom Log 

 Enter Choice
>>

tool: D27: %d

tool: D3: Enter Number

>>

tool: D61: Logarithm %d to the base 10 = %f

tool: D31: Logarithm %d to the base e = %f

tool: D38: Enter base

>>

tool: D52: Enter number

>>

assistant: ```c
void Logarithm() {
  int input, output;
  double i, x;
  printf("1-Compute log base 10 \n2- calculate natural logarithm \n3-Custom Log \n\n Enter Choice\n>>");
  scanf("%d", &input);
  if (input == 1) {
    printf("Enter Number\n\n>>");
    scanf("%d", &input);
    printf("Logarithm %d to the base 10 = %f", input, log10(input));
  } else if (input == 2) {
    printf("Enter Number\n\n>>");
    scanf("%d", &input);
    printf("Logarithm %d to the base e = %f", input, log(input));
  } else if (input == 3) {
    printf("Enter base\n\n>>");
    scanf("%d", &input);
    printf("Enter number\n\n>>");
    scanf("%d", &x);
    i = log(input) / log(x);
    printf("Logarithm %d to the base e = %f", input, log(i));
  }
}
```