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

```asm
<main>:
  endbr64
  push   %r15
  lea    D55(%rip),%rdi
  push   %r14
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  lea    D53(%rip),%rbp
  push   %rbx
  lea    D108(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D43(%rip),%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm4
  movd   %xmm4,%r14d
  jmp    L29
  nopl   0x0(%rax,%rax,1)
L52:
  pxor   %xmm2,%xmm2
  cvtss2sd %xmm0,%xmm2
L76:
  movss  0x4(%rsp),%xmm0
  mov    %r12d,%esi
  mov    $0x3,%eax
  lea    D24(%rip),%rdi
  add    $0x1,%r12d
  movaps %xmm0,%xmm1
  mulss  %xmm0,%xmm1
  subss  %xmm0,%xmm1
  subss  D99(%rip),%xmm1
  cvtss2sd %xmm0,%xmm0
  cvtss2sd %xmm1,%xmm1
  call   <printf@plt>
  mov    %r14d,0x4(%rsp)
  cmp    $0x1388,%r12d
  je     L11
  movd   %r15d,%xmm5
  pxor   %xmm0,%xmm0
  cvtss2sd %xmm5,%xmm0
  comisd D70(%rip),%xmm0
  jbe    L11
L29:
  movss  D99(%rip),%xmm0
  movd   %r14d,%xmm1
  divss  %xmm1,%xmm0
  addss  D90(%rip),%xmm0
  movd   %xmm0,%r14d
  subss  %xmm1,%xmm0
  call   <fabs@plt>
  movd   %xmm0,%r15d
  test   %r12d,%r12d
  jne    L52
  movss  0x4(%rsp),%xmm0
  movss  D52(%rip),%xmm1
  mulss  %xmm0,%xmm0
  divss  %xmm0,%xmm1
  comiss D90(%rip),%xmm1
  ja     L49
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    D54(%rip),%rax
  movq   %rax,%xmm2
  jmp    L76
  nopl   0x0(%rax,%rax,1)
L11:
  movd   %r14d,%xmm7
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  lea    D31(%rip),%rdi
  cvtss2sd %xmm7,%xmm0
  call   <printf@plt>
L54:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  pop    %r15
  ret
L49:
  lea    D123(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L54
L41:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D55
type: string
name: disassemble_data
arguments: label: D53
type: string
name: disassemble_data
arguments: label: D108
type: string
name: disassemble_data
arguments: label: D43
type: string
name: disassemble_data
arguments: label: D24
type: string
name: disassemble_data
arguments: label: D99
type: byte[4]
name: disassemble_data
arguments: label: D70
type: f64[2]
name: disassemble_data
arguments: label: D90
type: byte[4]
name: disassemble_data
arguments: label: D52
type: byte[4]
name: disassemble_data
arguments: label: D54
type: f32[2]
name: disassemble_data
arguments: label: D31
type: string
name: disassemble_data
arguments: label: D123
type: string

tool: D55: Please enter the initial value: 

tool: D53: 
   i	  xi	  f(xi)	         error

tool: D108: 
-------------------------------------------

tool: D43: %f

tool: D24: 
 %3d	%3.5f	 %3.5f	%3.5f

tool: {"D99": [0x00,0x00,0xc0,0x40]}

tool: {"D70": [0.0005,0.007812501884473022]}

tool: {"D90": [0x00,0x00,0x80,0x3f]}

tool: {"D52": [0x00,0x00,0xc0,0xc0]}

tool: {"D54": [0.0,-1.875]}

tool: D31: 
The required root is: %f


tool: D123: 
The equation is not convergent

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;
}
```