user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    D48(%rip),%rdi
  call   <printf@plt>
  lea    -0x3c(%rbp),%rsi
  lea    D23(%rip),%rdi
  call   <scanf@plt>
  movslq -0x3c(%rbp),%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L125:
  cmp    %rdx,%rsp
  je     L78
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L125
L78:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L135
  orq    $0x0,-0x8(%rsp,%rax,1)
L135:
  mov    %rsp,%rbx
  lea    D51(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x3c(%rbp),%eax
  test   %eax,%eax
  jle    L99
  mov    %rbx,%r13
  mov    $0x0,%r12d
  lea    D23(%rip),%r14
L127:
  mov    %r13,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  add    $0x1,%r12d
  mov    -0x3c(%rbp),%eax
  add    $0x4,%r13
  cmp    %r12d,%eax
  jg     L127
L99:
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L147:
  cmp    %rdx,%rsp
  je     L12
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L147
L12:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L74
  orq    $0x0,-0x8(%rsp,%rax,1)
L74:
  mov    %rsp,%r12
  lea    D60(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x0,-0x3c(%rbp)
  jle    L144
  mov    %r12,%r14
  mov    $0x0,%r13d
  lea    D23(%rip),%r15
L128:
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <scanf@plt>
  add    $0x1,%r13d
  mov    -0x3c(%rbp),%eax
  add    $0x4,%r14
  cmp    %r13d,%eax
  jg     L128
  test   %eax,%eax
  jle    L59
  mov    %eax,%ecx
  mov    $0x0,%eax
  mov    $0x0,%esi
L143:
  mov    (%rbx,%rax,4),%edx
  imul   (%r12,%rax,4),%edx
  add    %edx,%esi
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L143
L86:
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L81
  mov    $0x0,%eax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L144:
  mov    $0x0,%esi
  jmp    L86
L59:
  mov    $0x0,%esi
  jmp    L86
L81:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D48
type_guessed: string
name: parse_data
arguments: data_label: D23
type_guessed: string
name: parse_data
arguments: data_label: D51
type_guessed: string
name: parse_data
arguments: data_label: D60
type_guessed: string
name: parse_data
arguments: data_label: D2
type_guessed: string

tool: D48: Enter length of the vectors

tool: D23: %d

tool: D51: Enter values for vector 1: 

tool: D60: Enter values for vector 2: 

tool: D2: The dot product of the two vectors is %d

assistant: ```c
int main(void) {
  int i, length, dotProduct;
  printf("Enter length of the vectors");
  scanf("%d", &length);
  int vector1[length];
  printf("Enter values for vector 1: ");
  for (i = 0; i < length; ++i)
    scanf("%d", &vector1[i]);
  int vector2[length];
  printf("Enter values for vector 2: ");
  for (i = 0; i < length; ++i)
    scanf("%d", &vector2[i]);
  dotProduct = 0;
  for (i = 0; i < length; i++)
    dotProduct += vector1[i] * vector2[i];
  printf("The dot product of the two vectors is %d", dotProduct);
  return 0;
}
```