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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  movb   $0x0,-0x19(%rbp)
  lea    D131(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
L87:
  movss  -0x18(%rbp),%xmm2
  movss  D76(%rip),%xmm0
  movaps %xmm0,%xmm1
  divss  %xmm2,%xmm1
  movss  D37(%rip),%xmm0
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x10(%rbp)
  movss  -0x18(%rbp),%xmm1
  movss  -0x10(%rbp),%xmm0
  subss  %xmm1,%xmm0
  movd   %xmm0,%eax
  movd   %eax,%xmm0
  call   <fabs@plt>
  movd   %xmm0,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L61
  movss  -0x18(%rbp),%xmm1
  movss  -0x18(%rbp),%xmm0
  mulss  %xmm0,%xmm1
  movss  D32(%rip),%xmm0
  divss  %xmm1,%xmm0
  movss  D37(%rip),%xmm1
  comiss %xmm1,%xmm0
  jbe    L16
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movb   $0x1,-0x19(%rbp)
  jmp    L37
L16:
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D75(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L61:
  cmpl   $0x0,-0x14(%rbp)
  je     L79
  pxor   %xmm0,%xmm0
  cvtss2sd -0xc(%rbp),%xmm0
  jmp    L85
L79:
  movsd  D58(%rip),%xmm0
L85:
  movss  -0x18(%rbp),%xmm2
  movss  -0x18(%rbp),%xmm1
  mulss  %xmm2,%xmm1
  movss  -0x18(%rbp),%xmm2
  subss  %xmm2,%xmm1
  movss  D76(%rip),%xmm2
  subss  %xmm2,%xmm1
  cvtss2sd %xmm1,%xmm1
  movss  -0x18(%rbp),%xmm2
  pxor   %xmm3,%xmm3
  cvtss2sd %xmm2,%xmm3
  movq   %xmm3,%rdx
  mov    -0x14(%rbp),%eax
  movapd %xmm0,%xmm2
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  movss  -0x10(%rbp),%xmm0
  movss  %xmm0,-0x18(%rbp)
  addl   $0x1,-0x14(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0xc(%rbp),%xmm0
  comisd D28(%rip),%xmm0
  jbe    L37
  cmpl   $0x1387,-0x14(%rbp)
  jle    L87
L37:
  cmpb   $0x0,-0x19(%rbp)
  jne    L48
  movss  -0x18(%rbp),%xmm0
  pxor   %xmm4,%xmm4
  cvtss2sd %xmm0,%xmm4
  movq   %xmm4,%rax
  movq   %rax,%xmm0
  lea    D91(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
L48:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D131
data_type: string
name: parse_data
arguments: data_tag: D16
data_type: string
name: parse_data
arguments: data_tag: D76
data_type: byte[4]
name: parse_data
arguments: data_tag: D37
data_type: byte[4]
name: parse_data
arguments: data_tag: D32
data_type: byte[4]
name: parse_data
arguments: data_tag: D11
data_type: string
name: parse_data
arguments: data_tag: D26
data_type: string
name: parse_data
arguments: data_tag: D75
data_type: string
name: parse_data
arguments: data_tag: D58
data_type: f32[2]
name: parse_data
arguments: data_tag: D12
data_type: string
name: parse_data
arguments: data_tag: D28
data_type: f64
name: parse_data
arguments: data_tag: D91
data_type: string

tool: D131: Please enter the initial value: 

tool: D16: %f

tool: {"D76": [0x00,0x00,0xc0,0x40]}

tool: {"D37": [0x00,0x00,0x80,0x3f]}

tool: {"D32": [0x00,0x00,0xc0,0xc0]}

tool: D11: 
The equation is not convergent

tool: D26: 
   i	  xi	  f(xi)	         error

tool: D75: 
-------------------------------------------

tool: {"D58": [0.0,-1.875]}

tool: D12: 
 %3d	%3.5f	 %3.5f	%3.5f

tool: D28: 0.0005

tool: D91: 
The required root is: %f


assistant: ```c
int main() {
  int count = 0;
  float x0 = 0, x1 = 0, error = 0;
  char iffound = 0;
  printf("Please enter the initial value: ");
  scanf("%f", &x0);
  do {
    x1 = (1 + (6 / x0));
    error = fabs(x1 - x0);
    if (count == 0) {
      if (((-1) * 6 / (x0 * x0)) > 1) {
        printf("\nThe equation is not convergent");
        iffound = 1;
        break;
      } else {
        printf("\n   i	  xi	  f(xi)	         error");
        printf("\n-------------------------------------------");
      }
    }
    printf("\n %3d	%3.5f	 %3.5f	%3.5f", count, x0, (x0 * x0 - x0 - 6), count == 0 ? -1 : error);
    x0 = x1;
    count++;
  } while (error > 0.0005 && count < 5000);
  if (!iffound)
    printf("\nThe required root is: %f\n", x0);
  return 0;
}
```