user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movsd  D0(%rip),%xmm0
  movsd  %xmm0,-0x38(%rbp)
  movsd  D2(%rip),%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  D97(%rip),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movl   $0x0,-0x3c(%rbp)
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  call   <log@plt>
  movq   %xmm0,%rbx
  mov    -0x28(%rbp),%rax
  movq   %rax,%xmm0
  call   <log@plt>
  movq   %rbx,%xmm5
  divsd  %xmm0,%xmm5
  movsd  %xmm5,-0x48(%rbp)
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rcx
  mov    -0x3c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x3c(%rbp)
  movsd  -0x20(%rbp),%xmm0
  mov    -0x18(%rbp),%rdx
  movsd  -0x48(%rbp),%xmm3
  movapd %xmm0,%xmm2
  movq   %rcx,%xmm1
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x4,%eax
  call   <printf@plt>
L24:
  mov    -0x30(%rbp),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm4
  mulsd  -0x38(%rbp),%xmm4
  movsd  %xmm4,-0x48(%rbp)
  mov    -0x38(%rbp),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  mulsd  -0x30(%rbp),%xmm0
  movsd  -0x48(%rbp),%xmm4
  subsd  %xmm0,%xmm4
  movsd  %xmm4,-0x48(%rbp)
  mov    -0x30(%rbp),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rbx
  mov    -0x38(%rbp),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movq   %rbx,%xmm1
  subsd  %xmm0,%xmm1
  movsd  -0x48(%rbp),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  subsd  -0x30(%rbp),%xmm0
  movq   D7(%rip),%xmm1
  andpd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  call   <log@plt>
  movq   %xmm0,%rbx
  mov    -0x28(%rbp),%rax
  movq   %rax,%xmm0
  call   <log@plt>
  movq   %rbx,%xmm6
  divsd  %xmm0,%xmm6
  movsd  %xmm6,-0x48(%rbp)
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rcx
  mov    -0x3c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x3c(%rbp)
  movsd  -0x20(%rbp),%xmm0
  mov    -0x18(%rbp),%rdx
  movsd  -0x48(%rbp),%xmm3
  movapd %xmm0,%xmm2
  movq   %rcx,%xmm1
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D76(%rip),%rax
  mov    %rax,%rdi
  mov    $0x4,%eax
  call   <printf@plt>
  movsd  -0x30(%rbp),%xmm0
  movsd  %xmm0,-0x38(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rax
  pxor   %xmm0,%xmm0
  movq   %rax,%xmm7
  ucomisd %xmm0,%xmm7
  jp     L24
  pxor   %xmm0,%xmm0
  movq   %rax,%xmm2
  ucomisd %xmm0,%xmm2
  je     L39
  jmp    L24
L39:
  movsd  -0x18(%rbp),%xmm0
  subsd  -0x30(%rbp),%xmm0
  movq   D7(%rip),%xmm1
  andpd  %xmm1,%xmm0
  comisd D91(%rip),%xmm0
  ja     L24
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  lea    D39(%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: data_tag: D50
data_type: string
name: parse_data
arguments: data_tag: D38
data_type: string
name: parse_data
arguments: data_tag: D0
data_type: f64
name: parse_data
arguments: data_tag: D2
data_type: f64
name: parse_data
arguments: data_tag: D97
data_type: f64
name: parse_data
arguments: data_tag: D5
data_type: string
name: parse_data
arguments: data_tag: D7
data_type: qword
name: parse_data
arguments: data_tag: D76
data_type: string
name: parse_data
arguments: data_tag: D91
data_type: f64
name: parse_data
arguments: data_tag: D39
data_type: string

tool: D50: Solution of e^x = 2x +1 using secants method

tool: D38: Initial guess: 1

tool: D0: 1

tool: D2: 3

tool: D97: 2

tool: D5: i	Root		f(Root)		Error		order of convergence


tool: D7: 9223372036854775807

tool: D76: %d	%f	%f	%f	%f


tool: D91: 1e-06

tool: D39: 

Final Answer:%f


assistant: ```c
int main() {
  printf("Solution of e^x = 2x +1 using secants method\n");
  printf("Initial guess: 1\n");
  double veryOldRoot = 1, oldRoot = 3, newRoot;
  double oldError = 2;
  double newError;
  int i = 0;
  printf("i\tRoot\t\tf(Root)\t\tError\t\torder of convergence\n", i++, newRoot, f(newRoot), newError, log(newError) / log(oldError));
  do {
    newRoot = ((veryOldRoot * f(oldRoot) - oldRoot * f(veryOldRoot)) / (f(oldRoot) - f(veryOldRoot)));
    newError = fabs(newRoot - oldRoot);
    printf("%d\t%f\t%f\t%f\t%f\n", i++, newRoot, f(newRoot), newError, log(newError) / log(oldError));
    veryOldRoot = oldRoot;
    oldRoot = newRoot;
    oldError = newError;
  } while ((f(newRoot) != 0) || (fabs(newRoot - oldRoot) > 0.000001));
  printf("\n\nFinal Answer:%f\n", newRoot);
  return 0;
}
```