user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D68(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D11(%rip),%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%r12
  lea    0x20(%rsp),%r13
  call   <puts@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  lea    0x28(%rsp),%r14
  call   <scanf@plt>
  lea    D21(%rip),%rdi
  lea    0x30(%rsp),%r15
  call   <puts@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D124(%rip),%rdi
  call   <puts@plt>
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D73(%rip),%rdi
  call   <puts@plt>
  mov    %r15,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <calcAngularCoeff@plt>
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <calcIntercetta@plt>
  movsd  0x18(%rsp),%xmm3
  movsd  0x8(%rsp),%xmm2
  movapd %xmm0,%xmm1
  movsd  0x28(%rsp),%xmm0
  ucomisd %xmm0,%xmm3
  jp     L151
  movsd  D20(%rip),%xmm4
  jne    L151
L48:
  ucomisd 0x20(%rsp),%xmm4
  jp     L121
  movsd  D20(%rip),%xmm4
  jne    L121
L35:
  ucomisd 0x30(%rsp),%xmm4
  jp     L31
  je     L97
L31:
  subsd  %xmm0,%xmm3
  pxor   %xmm0,%xmm0
  ucomisd %xmm0,%xmm3
  jp     L127
  je     L106
L127:
  movsd  D20(%rip),%xmm3
  mov    $0x0,%edx
  ucomisd %xmm3,%xmm2
  setnp  %al
  cmovne %edx,%eax
  test   %al,%al
  je     L105
  ucomisd %xmm0,%xmm1
  setnp  %al
  cmovne %edx,%eax
  test   %al,%al
  jne    L27
L105:
  ucomisd %xmm0,%xmm2
  mov    $0x0,%edx
  setnp  %al
  cmovne %edx,%eax
  test   %al,%al
  je     L156
  ucomisd %xmm3,%xmm1
  setnp  %al
  cmovne %edx,%eax
  test   %al,%al
  je     L156
  lea    D36(%rip),%rdi
  call   <puts@plt>
L122:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L121:
  pxor   %xmm4,%xmm4
  jmp    L35
L151:
  pxor   %xmm4,%xmm4
  jmp    L48
L156:
  movapd %xmm2,%xmm0
  lea    D53(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  jmp    L122
L97:
  lea    D97(%rip),%rdi
  call   <puts@plt>
  jmp    L122
L27:
  lea    D80(%rip),%rdi
  call   <puts@plt>
  jmp    L122
L106:
  lea    D16(%rip),%rdi
  call   <puts@plt>
  jmp    L122
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D68
type: string
name: parse_data
arguments: label: D11
type: string
name: parse_data
arguments: label: D21
type: string
name: parse_data
arguments: label: D124
type: string
name: parse_data
arguments: label: D73
type: string
name: parse_data
arguments: label: D20
type: f32[2]
name: parse_data
arguments: label: D36
type: string
name: parse_data
arguments: label: D53
type: string
name: parse_data
arguments: label: D97
type: string
name: parse_data
arguments: label: D80
type: string
name: parse_data
arguments: label: D16
type: string

tool: D68: Enter the value of x1:

tool: D11: %lf

tool: D21: Enter the value of y1:

tool: D124: Enter the value of x2:

tool: D73: Enter the value of y2:

tool: {"D20": [0.0,1.875]}

tool: D36: The equation of the line is constant: y = 1.000

tool: D53: y = %lf x + %lf


tool: D97: The given points are equals.

tool: D80: The equation of the line goes through the origin: y = 1.000 * x

tool: D16: The line is vertical.

assistant: ```c
int main() {
  double x1, y1, x2, y2, m = 0, q = 0;
  printf("Enter the value of x1:\n");
  scanf("%lf", &x1);
  printf("Enter the value of y1:\n");
  scanf("%lf", &y1);
  printf("Enter the value of x2:\n");
  scanf("%lf", &x2);
  printf("Enter the value of y2:\n");
  scanf("%lf", &y2);
  m = calcAngularCoeff(&x1, &y1, &x2, &y2);
  q = calcIntercetta(&x1, &y1, &x2, &y2);
  if (x1 == x2 == y1 == y2) {
    printf("The given points are equals.\n");
    return 0;
  } else if ((x1 - x2) == 0) {
    printf("The line is vertical.\n");
    return 0;
  } else if ((m == 1) && (q == 0)) {
    printf("The equation of the line goes through the origin: y = 1.000 * x\n");
    return 0;
  } else if ((m == 0) && (q == 1)) {
    printf("The equation of the line is constant: y = 1.000\n");
    return 0;
  } else {
    printf("y = %lf x + %lf\n", m, q);
    return 0;
  }
  return 0;
}
```