user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x18(%rbp)
L12:
  lea    D66(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rdx
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1c(%rbp),%eax
  movd   %eax,%xmm0
  call   <f@plt>
  movsd  %xmm0,-0x28(%rbp)
  mov    -0x18(%rbp),%eax
  movd   %eax,%xmm0
  call   <f@plt>
  mulsd  -0x28(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm1,%xmm0
  ja     L12
L61:
  movss  -0x1c(%rbp),%xmm1
  movss  -0x18(%rbp),%xmm0
  addss  %xmm1,%xmm0
  movss  D23(%rip),%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movd   %eax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rcx
  pxor   %xmm0,%xmm0
  cvtss2sd -0x10(%rbp),%xmm0
  pxor   %xmm4,%xmm4
  cvtss2sd -0xc(%rbp),%xmm4
  movq   %xmm4,%rax
  mov    -0x14(%rbp),%edx
  movq   %rcx,%xmm2
  mov    %edx,%esi
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D61(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  movss  -0xc(%rbp),%xmm0
  ucomiss -0x10(%rbp),%xmm0
  jp     L50
  movss  -0xc(%rbp),%xmm0
  ucomiss -0x10(%rbp),%xmm0
  je     L77
L50:
  mov    -0x1c(%rbp),%eax
  movd   %eax,%xmm0
  call   <f@plt>
  movsd  %xmm0,-0x28(%rbp)
  mov    -0xc(%rbp),%eax
  movd   %eax,%xmm0
  call   <f@plt>
  movsd  -0x28(%rbp),%xmm1
  mulsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  comisd %xmm1,%xmm0
  jbe    L14
  movss  -0xc(%rbp),%xmm0
  movss  %xmm0,-0x18(%rbp)
  jmp    L63
L14:
  mov    -0xc(%rbp),%eax
  movd   %eax,%xmm0
  call   <f@plt>
  movsd  %xmm0,-0x28(%rbp)
  mov    -0x18(%rbp),%eax
  movd   %eax,%xmm0
  call   <f@plt>
  movsd  -0x28(%rbp),%xmm1
  mulsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  comisd %xmm1,%xmm0
  jbe    L63
  movss  -0xc(%rbp),%xmm0
  movss  %xmm0,-0x1c(%rbp)
L63:
  movss  -0xc(%rbp),%xmm0
  movss  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  movd   %eax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rax
  pxor   %xmm0,%xmm0
  movq   %rax,%xmm5
  ucomisd %xmm0,%xmm5
  jp     L61
  pxor   %xmm0,%xmm0
  movq   %rax,%xmm6
  ucomisd %xmm0,%xmm6
  je     L33
  jmp    L61
L77:
  nop
L33:
  movss  -0x18(%rbp),%xmm0
  cvtss2sd %xmm0,%xmm0
  movss  -0x1c(%rbp),%xmm1
  pxor   %xmm7,%xmm7
  cvtss2sd %xmm1,%xmm7
  movq   %xmm7,%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  pxor   %xmm2,%xmm2
  cvtss2sd -0xc(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D67(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L54
  call   <__stack_chk_fail@plt>
L54:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D66
type_guessed: string
name: unpack_data
arguments: label: D4
type_guessed: string
name: unpack_data
arguments: label: D23
type_guessed: byte[4]
name: unpack_data
arguments: label: D61
type_guessed: string
name: unpack_data
arguments: label: D14
type_guessed: string
name: unpack_data
arguments: label: D67
type_guessed: string
name: unpack_data
arguments: label: D40
type_guessed: string

tool: D66: Enter values of interval:

tool: D4: %f %f

tool: {"D23": [0x00,0x00,0x00,0x40]}

tool: D61: mid = %f newmid = %f i= %d and f(mid) = %f


tool: D14: The Interval is [%f, %f]


tool: D67: Root of f(x) is : %f


tool: D40: No of iterations: %d


assistant: ```c
void main() {
  int i = 0;
  float newmid = 0, mid = 0, a = 0, b = 0;
  do {
    printf("Enter values of interval:");
    scanf("%f %f", &a, &b);
  } while (f(a) * f(b) > 0);
  do {
    mid = (a + b) / 2;
    printf("mid = %f newmid = %f i= %d and f(mid) = %f\n", mid, newmid, i, f(mid));
    if (mid == newmid)
      break;
    else if (f(a) * f(mid) < 0)
      b = mid;
    else if (f(mid) * f(b) < 0)
      a = mid;
    newmid = mid;
    i++;
  } while (f(mid) != 0);
  printf("The Interval is [%f, %f]\n", a, b);
  printf("Root of f(x) is : %f\n", mid);
  printf("No of iterations: %d\n", i);
}
```