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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L32:
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D122(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x24(%rbp),%eax
  cmp    $0x3,%eax
  je     L107
  cmp    $0x3,%eax
  jg     L85
  cmp    $0x2,%eax
  je     L81
  cmp    $0x2,%eax
  jg     L85
  test   %eax,%eax
  je     L43
  cmp    $0x1,%eax
  jne    L85
  lea    D117(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D101(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D91(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D101(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movsd  -0x18(%rbp),%xmm0
  movsd  -0x20(%rbp),%xmm1
  subsd  %xmm1,%xmm0
  mov    -0x28(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x28(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%edi
  movq   %rax,%xmm0
  call   <Euller@plt>
  jmp    L85
L81:
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D101(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x28(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%edi
  movq   %rax,%xmm0
  call   <Euller@plt>
  jmp    L85
L107:
  mov    $0x0,%eax
  call   <main>
  jmp    L85
L43:
  mov    $0x0,%edi
  call   <exit@plt>
  nop
L85:
  mov    -0x24(%rbp),%eax
  test   %eax,%eax
  jne    L32
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L50
  call   <__stack_chk_fail@plt>
L50:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D30
type_infered: string
name: retrieve_data
arguments: tag: D39
type_infered: string
name: retrieve_data
arguments: tag: D41
type_infered: string
name: retrieve_data
arguments: tag: D7
type_infered: string
name: retrieve_data
arguments: tag: D122
type_infered: string
name: retrieve_data
arguments: tag: D106
type_infered: string
name: retrieve_data
arguments: tag: D5
type_infered: string
name: retrieve_data
arguments: tag: D117
type_infered: string
name: retrieve_data
arguments: tag: D101
type_infered: string
name: retrieve_data
arguments: tag: D91
type_infered: string
name: retrieve_data
arguments: tag: D34
type_infered: string
name: retrieve_data
arguments: tag: D18
type_infered: string

tool: D30:             Resolucao de Equacoes Diferenciais Pelo Metodo de Euller


tool: D39: 


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


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


tool: D7: 	(3)..........................Entrar Com Novos Valores de a e b


tool: D122: 	(0)..........................Sair


tool: D106: 
Opcao Desejada: 

tool: D5: %d

tool: D117: Entre com a: 

tool: D101: %lf

tool: D91: Entre com b: 

tool: D34: Entre com o numero de intervalos: 

tool: D18: 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;
}
```