user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r15
  lea    D90(%rip),%r15
  push   %r14
  lea    D53(%rip),%r14
  push   %r13
  lea    D41(%rip),%r13
  push   %r12
  lea    D10(%rip),%r12
  push   %rbp
  lea    D116(%rip),%rbp
  push   %rbx
  lea    D100(%rip),%rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x1c(%rsp),%rax
  mov    %rax,0x8(%rsp)
L31:
  mov    %r15,%rdi
  call   <printf@plt>
  mov    %r14,%rdi
  call   <printf@plt>
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    D141(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x1c(%rsp),%eax
  cmp    $0x2,%eax
  je     L86
  jg     L75
  test   %eax,%eax
  je     L21
  cmp    $0x1,%eax
  jne    L31
  lea    D73(%rip),%rdi
  call   <printf@plt>
  lea    0x20(%rsp),%rsi
  lea    D85(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D91(%rip),%rdi
  call   <printf@plt>
  lea    0x28(%rsp),%rsi
  lea    D85(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D61(%rip),%rdi
  call   <printf@plt>
  lea    0x18(%rsp),%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x18(%rsp),%edi
  pxor   %xmm1,%xmm1
  movsd  0x28(%rsp),%xmm0
  subsd  0x20(%rsp),%xmm0
  cvtsi2sd %edi,%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,0x30(%rsp)
  call   <Euller@plt>
  mov    0x1c(%rsp),%eax
L59:
  test   %eax,%eax
  jne    L31
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L75:
  cmp    $0x3,%eax
  jne    L31
  xor    %eax,%eax
  call   <main>
  mov    0x1c(%rsp),%eax
  jmp    L59
  cs nopw 0x0(%rax,%rax,1)
L21:
  xor    %edi,%edi
  call   <exit@plt>
  mov    0x1c(%rsp),%eax
  jmp    L59
  nopl   (%rax)
L86:
  lea    D50(%rip),%rdi
  call   <printf@plt>
  lea    0x30(%rsp),%rsi
  lea    D85(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D61(%rip),%rdi
  call   <printf@plt>
  lea    0x18(%rsp),%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x18(%rsp),%edi
  movsd  0x30(%rsp),%xmm0
  call   <Euller@plt>
  mov    0x1c(%rsp),%eax
  jmp    L59
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D90
type_infered: string
name: unpack_data
arguments: data_label: D53
type_infered: string
name: unpack_data
arguments: data_label: D41
type_infered: string
name: unpack_data
arguments: data_label: D10
type_infered: string
name: unpack_data
arguments: data_label: D116
type_infered: string
name: unpack_data
arguments: data_label: D100
type_infered: string
name: unpack_data
arguments: data_label: D141
type_infered: string
name: unpack_data
arguments: data_label: D73
type_infered: string
name: unpack_data
arguments: data_label: D85
type_infered: string
name: unpack_data
arguments: data_label: D91
type_infered: string
name: unpack_data
arguments: data_label: D61
type_infered: string
name: unpack_data
arguments: data_label: D50
type_infered: string

tool: D90:             Resolucao de Equacoes Diferenciais Pelo Metodo de Euller


tool: D53: 


	(1)..........................Entrar com a, b e n


tool: D41: 	(2)..........................Entrar com h e n


tool: D10: 	(3)..........................Entrar Com Novos Valores de a e b


tool: D116: 	(0)..........................Sair


tool: D100: %d

tool: D141: 
Opcao Desejada: 

tool: D73: Entre com a: 

tool: D85: %lf

tool: D91: Entre com b: 

tool: D61: Entre com o numero de intervalos: 

tool: D50: Entre com h: 

assistant: ```c
int main() {
  double a, b, h;
  int n, opcao;
  do {
    printf("            Resolucao de Equacoes Diferenciais Pelo Metodo de Euller\n");
    printf("\n\n\n\t(1)..........................Entrar com a, b e n\n");
    printf("\t(2)..........................Entrar com h e n\n");
    printf("\t(3)..........................Entrar Com Novos Valores de a e b\n");
    printf("\t(0)..........................Sair\n");
    printf("\nOpcao Desejada: ");
    scanf("%d", &opcao);
    switch (opcao) {
    case 1:
      printf("Entre com a: ");
      scanf("%lf", &a);
      printf("Entre com b: ");
      scanf("%lf", &b);
      printf("Entre com o numero de intervalos: ");
      scanf("%d", &n);
      h = (b - a) / n;
      Euller(h, n);
      break;
    case 2:
      printf("Entre com h: ");
      scanf("%lf", &h);
      printf("Entre com o numero de intervalos: ");
      scanf("%d", &n);
      Euller(h, n);
      break;
    case 3:
      main();
      break;
    case 0:
      exit(0);
      break;
    }
  } while (opcao);
  return 0;
}
```