user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<kmemmem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %esi,-0x4c(%rbp)
  mov    %rdx,-0x58(%rbp)
  mov    %ecx,-0x50(%rbp)
  mov    %r8,-0x60(%rbp)
  movq   $0x0,-0x28(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x58(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x60(%rbp)
  je     L144
  mov    -0x60(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L22
L144:
  mov    -0x50(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <ksBM_prep@plt>
  jmp    L46
L22:
  mov    -0x60(%rbp),%rax
  mov    (%rax),%rax
L46:
  mov    %rax,-0x28(%rbp)
  cmpq   $0x0,-0x60(%rbp)
  je     L97
  mov    -0x60(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L97
  mov    -0x60(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rdx,(%rax)
L97:
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x50(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x30(%rbp)
  jmp    L82
L30:
  mov    -0x50(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x34(%rbp)
  jmp    L4
L75:
  subl   $0x1,-0x34(%rbp)
L4:
  cmpl   $0x0,-0x34(%rbp)
  js     L119
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x34(%rbp),%ecx
  mov    -0x30(%rbp),%eax
  add    %ecx,%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L75
L119:
  cmpl   $0x0,-0x34(%rbp)
  js     L21
  mov    -0x34(%rbp),%edx
  mov    -0x30(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sub    -0x50(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x34(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x2c(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x2c(%rbp)
  jge    L55
  mov    -0x34(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x2c(%rbp)
L55:
  mov    -0x2c(%rbp),%eax
  add    %eax,-0x30(%rbp)
  jmp    L82
L21:
  mov    -0x30(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  jmp    L124
L82:
  mov    -0x4c(%rbp),%eax
  sub    -0x50(%rbp),%eax
  cmp    %eax,-0x30(%rbp)
  jle    L30
  cmpq   $0x0,-0x60(%rbp)
  jne    L107
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L107:
  mov    $0x0,%eax
L124:
  leave
  ret
```

assistant: ```c
void *kmemmem(const void *_str, int n, const void *_pat, int m, int **_prep) {
  int i, j;
  int *prep = 0;
  int *bmGs;
  int *bmBc;
  const unsigned char *str;
  const unsigned char *pat;
  str = (const unsigned char *)_str;
  pat = (const unsigned char *)_pat;
  prep = ((_prep == 0) || (*_prep == 0)) ? ksBM_prep(pat, m) : *_prep;
  if (_prep && (*_prep == 0))
    *_prep = prep;
  bmGs = prep;
  bmBc = prep + m;
  j = 0;
  while (j <= (n - m)) {
    for (i = m - 1; (i >= 0) && (pat[i] == str[i + j]); --i)
      ;
    if (i >= 0) {
      int max = bmBc[str[i + j]] - m + 1 + i;
      if (max < bmGs[i])
        max = bmGs[i];
      j += max;
    } else
      return (void *)(str + j);
  }
  if (_prep == 0)
    free(prep);
  return 0;
}
```