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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movsd  D39(%rip),%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  D49(%rip),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movl   $0x0,-0x34(%rbp)
  lea    D69(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L5:
  mov    -0x30(%rbp),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rbx
  mov    -0x30(%rbp),%rax
  movq   %rax,%xmm0
  call   <diff@plt>
  movq   %rbx,%xmm1
  divsd  %xmm0,%xmm1
  movsd  -0x30(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  subsd  -0x30(%rbp),%xmm0
  movq   D52(%rip),%xmm1
  andpd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <log@plt>
  movq   %xmm0,%rbx
  mov    -0x28(%rbp),%rax
  movq   %rax,%xmm0
  call   <log@plt>
  movq   %rbx,%xmm4
  divsd  %xmm0,%xmm4
  movsd  %xmm4,-0x48(%rbp)
  mov    -0x30(%rbp),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rcx
  mov    -0x34(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x34(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mov    -0x30(%rbp),%rdx
  movsd  -0x48(%rbp),%xmm3
  movapd %xmm0,%xmm2
  movq   %rcx,%xmm1
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D72(%rip),%rax
  mov    %rax,%rdi
  mov    $0x4,%eax
  call   <printf@plt>
  movsd  -0x30(%rbp),%xmm0
  subsd  -0x20(%rbp),%xmm0
  movq   D52(%rip),%xmm1
  andpd  %xmm0,%xmm1
  movsd  D95(%rip),%xmm0
  comisd %xmm1,%xmm0
  ja     L17
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  jmp    L5
L17:
  nop
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D44
data_type: string
name: parse_data
arguments: tag: D25
data_type: string
name: parse_data
arguments: tag: D39
data_type: f64
name: parse_data
arguments: tag: D49
data_type: f64
name: parse_data
arguments: tag: D69
data_type: string
name: parse_data
arguments: tag: D52
data_type: qword
name: parse_data
arguments: tag: D72
data_type: string
name: parse_data
arguments: tag: D95
data_type: f64
name: parse_data
arguments: tag: D40
data_type: string

tool: D44: Solution for e^x = 2x+1 using newton raphson method

tool: D25: Initial approximation:1

tool: D39: 1

tool: D49: 2

tool: D69: i	Root		f(Root)		Error		Order of convergence

tool: D52: 9223372036854775807

tool: D72: %d	%f	%f	%f	%f


tool: D95: 1e-06

tool: D40: 

Final Answer:%f


assistant: ```c
int main() {
  printf("Solution for e^x = 2x+1 using newton raphson method\n");
  printf("Initial approximation:1\n");
  double newRoot, oldRoot = 1;
  double oldError = 2, newError;
  int i = 0;
  printf("i\tRoot\t\tf(Root)\t\tError\t\tOrder of convergence\n");
  while (1) {
    newRoot = oldRoot - (f(oldRoot) / diff(oldRoot));
    newError = fabs(newRoot - oldRoot);
    printf("%d\t%f\t%f\t%f\t%f\n", i++, oldRoot, f(oldRoot), newError, log(newError) / log(oldError));
    if (fabs(oldRoot - newRoot) < 0.000001)
      break;
    oldRoot = newRoot;
    oldError = newError;
  }
  printf("\n\nFinal Answer:%f\n", newRoot);
  return 0;
}
```