user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<mine>:
  endbr64
  push   %r15
  lea    D30(%rip),%rdi
  push   %r14
  push   %r13
  lea    D7(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x1a8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x198(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r15
  call   <puts@plt>
  mov    %r15,%rbx
  nopl   0x0(%rax)
L63:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  mov    %r12d,%ebp
  add    $0x4,%rbx
  call   <scanf@plt>
  lea    0x1(%r12),%r12d
  cmp    $0x1,%eax
  je     L63
  lea    -0x3(%rbp),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L93
  mov    %ebp,(%rax)
  cmp    $0x1,%r12d
  je     L89
  sub    %r15,%rax
  cmp    $0x8,%rax
  jbe    L45
  cmp    $0x3,%ebp
  jle    L45
  test   %ebp,%ebp
  mov    $0x1,%ecx
  cmovg  %ebp,%ecx
  xor    %eax,%eax
  mov    %ecx,%edx
  shr    $0x2,%edx
  shl    $0x4,%rdx
  nop
L16:
  movdqa (%r15,%rax,1),%xmm0
  movups %xmm0,0x4(%r13,%rax,1)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L16
  mov    %ecx,%eax
  and    $0xfffffffc,%eax
  lea    0x1(%rax),%esi
  cmp    %eax,%ecx
  je     L48
  movslq %eax,%rdi
  movslq %esi,%rdx
  add    $0x2,%eax
  mov    (%rsp,%rdi,4),%edi
  lea    0x0(,%rdx,4),%rcx
  mov    %edi,0x0(%r13,%rdx,4)
  cmp    %esi,%ebp
  jle    L48
  mov    (%rsp,%rdx,4),%edx
  mov    %edx,0x4(%r13,%rcx,1)
  cmp    %eax,%ebp
  jle    L48
  cltq
  mov    (%rsp,%rax,4),%eax
  mov    %eax,0x8(%r13,%rcx,1)
L48:
  test   %ebp,%ebp
  mov    $0x1,%r14d
  cmovg  %ebp,%r14d
  add    $0x1,%r14d
L93:
  lea    D63(%rip),%rdi
  lea    0x4(%r13),%rbx
  call   <puts@plt>
  mov    0x0(%r13),%esi
  lea    D5(%rip),%rdi
  lea    D0(%rip),%rbp
  call   <printf@plt>
  lea    -0x2(%r14),%eax
  lea    0x8(%r13,%rax,4),%r12
  nopl   0x0(%rax)
L15:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L15
L134:
  lea    D6(%rip),%rdi
  call   <puts@plt>
  mov    0x198(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0x1a8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L45:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L37:
  mov    (%r15,%rax,4),%edx
  mov    %edx,0x4(%r13,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%ebp
  jg     L37
  jmp    L48
L89:
  lea    D63(%rip),%rdi
  call   <puts@plt>
  mov    0x0(%r13),%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  jmp    L134
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D30
type_guessed: string
name: parse_data
arguments: data_tag: D7
type_guessed: string
name: parse_data
arguments: data_tag: D63
type_guessed: string
name: parse_data
arguments: data_tag: D5
type_guessed: string
name: parse_data
arguments: data_tag: D0
type_guessed: string
name: parse_data
arguments: data_tag: D6
type_guessed: string

tool: D30: Enter some numbers separated by whitespaces:

tool: D7: %d

tool: D63: Your numbers, in a dynamic allocated int array would be..

tool: D5: %d: 

tool: D0: %d, 

tool: D6: 


assistant: ```c
void mine(void) {
  int array[100];
  int y, x = 0;
  puts("Enter some numbers separated by whitespaces:");
  while (scanf("%d", &array[x++]) == 1)
    ;
  int *ptr = malloc(x - 1 * sizeof(int));
  if (ptr != NULL) {
    ptr[0] = x - 1;
    for (y = 1; y < x; y++)
      ptr[y] = array[y - 1];
  }
  puts("Your numbers, in a dynamic allocated int array would be..");
  printf("%d: ", ptr[0]);
  x = 1;
  while (x < y) {
    printf("%d, ", ptr[x++]);
  }
  puts("\n");
}
```