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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x40(%rbp)
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x4c(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x44(%rbp)
  jmp    L87
L4:
  lea    -0x30(%rbp),%rdx
  mov    -0x44(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x44(%rbp)
L87:
  mov    -0x4c(%rbp),%eax
  cmp    %eax,-0x44(%rbp)
  jl     L4
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x48(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x3c(%rbp)
  mov    -0x4c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x38(%rbp)
  jmp    L67
L38:
  mov    -0x3c(%rbp),%edx
  mov    -0x38(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x48(%rbp),%eax
  cmp    %eax,%edx
  jne    L75
  mov    -0x34(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x48(%rbp),%eax
  mov    %eax,%esi
  lea    D75(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x40(%rbp)
  jmp    L43
L75:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x48(%rbp),%eax
  cmp    %eax,%edx
  jle    L13
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x38(%rbp)
  jmp    L67
L13:
  mov    -0x34(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x3c(%rbp)
L67:
  mov    -0x3c(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jle    L38
L43:
  cmpl   $0x0,-0x40(%rbp)
  je     L7
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L7:
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L40
  call   <__stack_chk_fail@plt>
L40:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D62
data_type: string
name: extract_symbolic_data
arguments: data_tag: D9
data_type: string
name: extract_symbolic_data
arguments: data_tag: D65
data_type: string
name: extract_symbolic_data
arguments: data_tag: D19
data_type: string
name: extract_symbolic_data
arguments: data_tag: D75
data_type: string
name: extract_symbolic_data
arguments: data_tag: D2
data_type: string

tool: D62: Enter the how mny nos......

tool: D9: %d

tool: D65: Enter the element in sorted order .......

tool: D19: Enter the searching element......

tool: D75: %d is in %d position


tool: D2: Not found

assistant: ```c
void main() {
  int i, a[10], n, src, flg = 1, lb, ub, mid;
  printf("Enter the how mny nos......\n");
  scanf("%d", &n);
  printf("Enter the element in sorted order .......\n");
  for (i = 0; i < n; i++)
    scanf("%d", &a[i]);
  printf("Enter the searching element......\n");
  scanf("%d", &src);
  lb = 0;
  ub = n - 1;
  while (lb <= ub) {
    mid = (lb + ub) / 2;
    if (a[mid] == src) {
      printf("%d is in %d position\n", src, mid + 1);
      flg = 0;
      break;
    } else if (src < a[mid])
      ub = mid - 1;
    else
      lb = mid + 1;
  }
  if (flg)
    printf("Not found");
  printf("\n");
}
```