user: 请从给定的汇编指令推导出C源代码：

```asm
<alloc_parse_doubles>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x68,%rsp
  mov    %edi,-0x74(%rbp)
  mov    %rsi,-0x80(%rbp)
  mov    %rdx,-0x88(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x80(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x5c(%rbp)
  mov    -0x5c(%rbp),%eax
  add    $0x2,%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x50(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  movslq %eax,%rdx
  mov    %rdx,%r12
  mov    $0x0,%r13d
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L86:
  cmp    %rdx,%rsp
  je     L35
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L86
L35:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L70
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L70:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x48(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x58(%rbp)
  mov    -0x58(%rbp),%rax
  movb   $0x20,(%rax)
  movl   $0x0,-0x64(%rbp)
  jmp    L68
L3:
  mov    -0x64(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x80(%rbp),%rax
  add    %rdx,%rax
  mov    -0x64(%rbp),%edx
  movslq %edx,%rdx
  lea    0x1(%rdx),%rcx
  mov    -0x58(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x64(%rbp)
L68:
  mov    -0x64(%rbp),%eax
  cmp    -0x5c(%rbp),%eax
  jle    L3
  mov    -0x74(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x40(%rbp)
  movl   $0x0,-0x60(%rbp)
  jmp    L63
L26:
  addl   $0x1,-0x60(%rbp)
  mov    -0x68(%rbp),%eax
  cltq
  add    %rax,-0x58(%rbp)
L63:
  mov    -0x60(%rbp),%eax
  cmp    -0x74(%rbp),%eax
  jge    L34
  mov    -0x60(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  lea    -0x68(%rbp),%rdx
  mov    -0x58(%rbp),%rax
  mov    %rdx,%rcx
  mov    %rsi,%rdx
  lea    D9(%rip),%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  je     L26
L34:
  mov    -0x88(%rbp),%rax
  mov    -0x60(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x40(%rbp),%rax
  mov    %rbx,%rsp
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D9
type: string

tool: D9: %*[][ 
	,:;]%lf%n

assistant: ```c
double *alloc_parse_doubles(int nmax, const char *ss, int *n) {
  int ns = strlen(ss);
  char t[2 + ns], *s = t;
  s[0] = ' ';
  for (int i = 0; i <= ns; i++)
    s[i + 1] = ss[i];
  double *r = (double *)malloc(nmax * sizeof *r);
  int i = 0, w;
  while (i < nmax && 1 == sscanf(s, "%*[][ \n\t,:;]%lf%n", r + i, &w)) {
    i += 1;
    s += w;
  }
  *n = i;
  return r;
}
```