user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<str_parse>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmp    $0x2,%eax
  jle    L99
  cmpb   $0x30,0x0(%r13)
  je     L37
L99:
  movslq %ebx,%rbx
  jmp    L214
  nopl   (%rax)
L116:
  add    $0x1,%rbx
L214:
  movsbl 0x0(%r13,%rbx,1),%edi
  mov    %ebx,%ebp
  mov    %ebx,%r12d
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L116
  movzbl 0x0(%r13,%rbx,1),%eax
  sub    $0x61,%eax
  cmp    $0x5,%al
  jbe    L116
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  test   %ebx,%ebx
  je     L16
  lea    0x1(%r13),%rcx
  mov    %r8,%rdx
  lea    -0x1(%rbx),%eax
  sub    %rcx,%rdx
  cmp    $0xe,%rdx
  jbe    L110
  cmp    $0x6,%eax
  jbe    L110
  cmp    $0xe,%eax
  jbe    L111
  mov    %ebx,%edx
  xor    %eax,%eax
  shr    $0x4,%edx
  shl    $0x4,%rdx
  nopl   (%rax)
L63:
  movdqu 0x0(%r13,%rax,1),%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L63
  mov    %ebp,%edx
  and    $0xfffffff0,%edx
  mov    %edx,%eax
  cmp    %edx,%ebp
  je     L16
  sub    %edx,%ebp
  lea    -0x1(%rbp),%ecx
  cmp    $0x6,%ecx
  jbe    L178
L90:
  mov    0x0(%r13,%rdx,1),%rcx
  mov    %rcx,(%r8,%rdx,1)
  mov    %ebp,%edx
  and    $0xfffffff8,%edx
  add    %edx,%eax
  cmp    %edx,%ebp
  je     L16
L178:
  movslq %eax,%rdx
  movzbl 0x0(%r13,%rdx,1),%ecx
  mov    %cl,(%r8,%rdx,1)
  lea    0x1(%rax),%edx
  cmp    %edx,%r12d
  jle    L16
  movslq %edx,%rdx
  movzbl 0x0(%r13,%rdx,1),%ecx
  mov    %cl,(%r8,%rdx,1)
  lea    0x2(%rax),%edx
  cmp    %edx,%r12d
  jle    L16
  movslq %edx,%rdx
  movzbl 0x0(%r13,%rdx,1),%ecx
  mov    %cl,(%r8,%rdx,1)
  lea    0x3(%rax),%edx
  cmp    %edx,%r12d
  jle    L16
  movslq %edx,%rdx
  movzbl 0x0(%r13,%rdx,1),%ecx
  mov    %cl,(%r8,%rdx,1)
  lea    0x4(%rax),%edx
  cmp    %edx,%r12d
  jle    L16
  movslq %edx,%rdx
  movzbl 0x0(%r13,%rdx,1),%ecx
  mov    %cl,(%r8,%rdx,1)
  lea    0x5(%rax),%edx
  cmp    %edx,%r12d
  jle    L16
  movslq %edx,%rdx
  add    $0x6,%eax
  movzbl 0x0(%r13,%rdx,1),%ecx
  mov    %cl,(%r8,%rdx,1)
  cmp    %eax,%r12d
  jle    L16
  cltq
  movzbl 0x0(%r13,%rax,1),%edx
  mov    %dl,(%r8,%rax,1)
L16:
  movb   $0x0,(%r8,%rbx,1)
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L37:
  xor    %ebx,%ebx
  cmpb   $0x78,0x1(%r13)
  sete   %bl
  add    %ebx,%ebx
  jmp    L99
L110:
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L156:
  movzbl 0x0(%r13,%rax,1),%edx
  mov    %dl,(%r8,%rax,1)
  add    $0x1,%rax
  cmp    %rbx,%rax
  jne    L156
  jmp    L16
L111:
  xor    %edx,%edx
  xor    %eax,%eax
  jmp    L90
```

assistant: ```c
char *str_parse(char *input) {
  int len = 0;
  if (strlen(input) > 2 && input[0] == '0' && input[1] == 'x') {
    len = 2;
  }
  while (isdigit(input[len]) || input[len] == 'a' || input[len] == 'b' || input[len] == 'c' || input[len] == 'd' || input[len] == 'e' || input[len] == 'f') {
    len += 1;
  }
  len += 1;
  char *output = malloc(sizeof(char) * len);
  for (int i = 0; i < (len - 1); i += 1) {
    output[i] = input[i];
  }
  output[len - 1] = '\0';
  return output;
}
```