user: 这段汇编代码用C语言表示是什么样的?

```asm
<rle>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  test   %edx,%edx
  jle    L124
  mov    %rsi,%r12
  mov    %rdi,%rbx
  mov    %edx,%esi
  xor    %r11d,%r11d
  mov    $0x2,%r13d
  cs nopw 0x0(%rax,%rax,1)
L58:
  movslq %r11d,%r10
  lea    0x1(%r11),%eax
  movzbl (%r12,%r10,1),%edi
  cmp    %eax,%esi
  jle    L194
  movslq %eax,%rdx
  mov    %r13d,%r8d
  lea    0x7f(%r11),%r9d
  add    %r12,%rdx
  sub    %edx,%r8d
  jmp    L206
  cs nopw 0x0(%rax,%rax,1)
L155:
  add    $0x1,%eax
  lea    (%r8,%rdx,1),%ecx
  cmp    %eax,%esi
  je     L191
  add    $0x1,%rdx
  cmp    %r9d,%eax
  je     L112
L206:
  movzbl (%rdx),%ecx
  cmp    %dil,%cl
  je     L155
  mov    %eax,%edx
  sub    %r11d,%edx
  cmp    $0x1,%edx
  je     L153
L66:
  mov    %eax,%edx
  mov    %dil,0x1(%rbx)
  add    $0x2,%rbx
  sub    %r11d,%edx
  mov    %eax,%r11d
  mov    %dl,-0x2(%rbx)
L172:
  cmp    %esi,%r11d
  jl     L58
  lea    0x1(%rbx),%rax
  movb   $0x0,(%rbx)
  pop    %rbx
  sub    %ebp,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L191:
  cmp    $0x1,%ecx
  jne    L66
  mov    %esi,%eax
  nop
L194:
  mov    %r11d,%edx
  lea    0x1(%rbx),%rcx
  sub    %eax,%edx
  mov    %dl,(%rbx)
  cmp    %r11d,%eax
  jle    L43
L115:
  sub    %r10,%rbx
  nopw   0x0(%rax,%rax,1)
L152:
  movzbl (%r12,%r10,1),%edx
  mov    %dl,0x1(%rbx,%r10,1)
  add    $0x1,%r10
  cmp    %r10d,%eax
  jg     L152
  mov    %eax,%edx
  sub    %r11d,%edx
  mov    %eax,%r11d
  sub    $0x1,%edx
  lea    0x1(%rcx,%rdx,1),%rbx
  jmp    L172
  nopl   (%rax)
L153:
  cmp    %eax,%esi
  jle    L194
  movslq %eax,%rdx
L24:
  mov    %eax,%edi
  add    %r12,%rdx
  sub    %r11d,%edi
  sub    %edx,%edi
  jmp    L142
  nopl   0x0(%rax,%rax,1)
L28:
  add    $0x1,%eax
  cmp    %eax,%esi
  jle    L194
  add    $0x1,%rdx
  lea    (%rdi,%rdx,1),%ecx
  cmp    $0x7e,%ecx
  jg     L194
  movzbl (%rdx),%ecx
L142:
  cmp    %cl,-0x1(%rdx)
  jne    L28
  cmp    $0x1,%eax
  jle    L28
  cmp    %cl,-0x2(%rdx)
  jne    L28
  sub    $0x2,%eax
  mov    %r11d,%edx
  lea    0x1(%rbx),%rcx
  sub    %eax,%edx
  mov    %dl,(%rbx)
  cmp    %r11d,%eax
  jg     L115
L43:
  mov    %rcx,%rbx
  mov    %eax,%r11d
  jmp    L172
L112:
  cmp    $0x1,%ecx
  jne    L66
  movslq %eax,%rdx
  movzbl (%r12,%rdx,1),%ecx
  jmp    L24
L124:
  mov    %rdi,%rbx
  lea    0x1(%rbx),%rax
  movb   $0x0,(%rbx)
  pop    %rbx
  sub    %ebp,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int rle(unsigned char *output, const unsigned char *input, int len) {
  int count, index, i;
  unsigned char first;
  unsigned char *out;
  out = output;
  count = 0;
  while (count < len) {
    index = count;
    first = input[index++];
    while ((index < len) && (index - count < 127) && (input[index] == first))
      index++;
    if (index - count == 1) {
      while ((index < len) && (index - count < 127)) {
        if ((input[index] == input[index - 1]) && ((index > 1) && (input[index] == input[index - 2]))) {
          index -= 2;
          break;
        }
        index++;
      }
      *out++ = (unsigned char)(count - index);
      for (i = count; i < index; i++)
        *out++ = input[i];
    } else {
      *out++ = (unsigned char)(index - count);
      *out++ = first;
    }
    count = index;
  }
  *out++ = 0;
  return (out - output);
}
```