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    $0x378,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movsd  D36(%rip),%xmm0
  movsd  %xmm0,-0x340(%rbp)
  movsd  D16(%rip),%xmm0
  movsd  %xmm0,-0x360(%rbp)
  movsd  D36(%rip),%xmm0
  movsd  %xmm0,-0x358(%rbp)
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x1,-0x36c(%rbp)
  jmp    L24
L2:
  mov    -0x36c(%rbp),%eax
  sub    $0x1,%eax
  cltq
  movsd  -0x340(%rbp,%rax,8),%xmm2
  movsd  %xmm2,-0x378(%rbp)
  mov    -0x36c(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x340(%rbp,%rax,8),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rbx
  mov    -0x36c(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x340(%rbp,%rax,8),%rax
  movq   %rax,%xmm0
  call   <fp@plt>
  movq   %rbx,%xmm1
  divsd  %xmm0,%xmm1
  movsd  -0x378(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  mov    -0x36c(%rbp),%eax
  cltq
  movsd  %xmm0,-0x340(%rbp,%rax,8)
  mov    -0x36c(%rbp),%eax
  cltq
  mov    -0x340(%rbp,%rax,8),%rdx
  mov    -0x36c(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    -0x36c(%rbp),%eax
  cltq
  movsd  -0x340(%rbp,%rax,8),%xmm0
  movsd  %xmm0,-0x368(%rbp)
  addl   $0x1,-0x36c(%rbp)
L24:
  cmpl   $0x64,-0x36c(%rbp)
  jle    L2
  mov    -0x360(%rbp),%rax
  movq   %rax,%xmm0
  call   <fp@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x350(%rbp)
  mov    -0x368(%rbp),%rax
  movq   %rax,%xmm0
  call   <f@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  divsd  -0x350(%rbp),%xmm0
  movsd  %xmm0,-0x348(%rbp)
  mov    -0x348(%rbp),%rax
  movq   %rax,%xmm0
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D36
type_infered: f64
name: read_data
arguments: label: D16
type_infered: f64
name: read_data
arguments: label: D10
type_infered: string
name: read_data
arguments: label: D45
type_infered: string
name: read_data
arguments: label: D5
type_infered: string
name: read_data
arguments: label: D27
type_infered: string
name: read_data
arguments: label: D3
type_infered: string

tool: D36: 3

tool: D16: 2

tool: D10: ****************************************************************

tool: D45: *  Estimacion de una solucion por el metodo de Newton-Raphson  *

tool: D5: iteracion %d
 estimacion:%.15lf


tool: D27: error= %.18lf


tool: D3: PAUSE

assistant: ```c
int main() {
  double x[101];
  int i;
  x[0] = 3;
  double a = 2;
  double b = 3;
  double error;
  double s;
  double minFp;
  printf("****************************************************************\n");
  printf("*  Estimacion de una solucion por el metodo de Newton-Raphson  *\n");
  printf("****************************************************************\n");
  for (i = 1; i <= 100; i++) {
    x[i] = x[i - 1] - (f(x[i - 1]) / fp(x[i - 1]));
    printf("iteracion %d\n estimacion:%.15lf\n", i, x[i]);
    s = x[i];
  }
  minFp = fp(a);
  error = (f(s)) / minFp;
  printf("error= %.18lf\n", error);
  system("PAUSE");
}
```