user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Base64Decode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %rdi,-0x88(%rbp)
  mov    %rsi,-0x90(%rbp)
  mov    %edx,-0x94(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x60(%rbp)
  movl   $0x0,-0x55(%rbp)
  movb   $0x0,-0x51(%rbp)
  movabs $0x4847464544434241,%rax
  movabs $0x504f4e4d4c4b4a49,%rdx
  mov    %rax,-0x50(%rbp)
  mov    %rdx,-0x48(%rbp)
  movabs $0x5857565554535251,%rax
  movabs $0x6665646362615a59,%rdx
  mov    %rax,-0x40(%rbp)
  mov    %rdx,-0x38(%rbp)
  movabs $0x6e6d6c6b6a696867,%rax
  movabs $0x767574737271706f,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movabs $0x333231307a797877,%rax
  movabs $0x2f2b393837363534,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movl   $0x0,-0x74(%rbp)
  movl   $0x0,-0x68(%rbp)
  movl   $0x0,-0x70(%rbp)
  movl   $0x0,-0x64(%rbp)
  movl   $0x0,-0x6c(%rbp)
  mov    -0x88(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x64(%rbp)
  jmp    L0
L27:
  mov    -0x74(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x88(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x68(%rbp)
  cmpl   $0x3d,-0x68(%rbp)
  jne    L6
  mov    -0x94(%rbp),%edx
  mov    -0x90(%rbp),%rcx
  lea    -0x55(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <decodeblock@plt>
  mov    %eax,-0x6c(%rbp)
  jmp    L63
L6:
  mov    -0x68(%rbp),%edx
  lea    -0x50(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x60(%rbp)
  cmpq   $0x0,-0x60(%rbp)
  je     L57
  lea    -0x50(%rbp),%rdx
  mov    -0x60(%rbp),%rax
  sub    %rdx,%rax
  mov    %eax,%edx
  mov    -0x70(%rbp),%eax
  cltq
  mov    %dl,-0x55(%rbp,%rax,1)
  mov    -0x70(%rbp),%eax
  add    $0x1,%eax
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,-0x70(%rbp)
  cmpl   $0x0,-0x70(%rbp)
  jne    L57
  mov    -0x94(%rbp),%edx
  mov    -0x90(%rbp),%rcx
  lea    -0x55(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <decodeblock@plt>
  mov    %eax,-0x6c(%rbp)
  movb   $0x0,-0x52(%rbp)
  movzbl -0x52(%rbp),%eax
  mov    %al,-0x53(%rbp)
  movzbl -0x53(%rbp),%eax
  mov    %al,-0x54(%rbp)
  movzbl -0x54(%rbp),%eax
  mov    %al,-0x55(%rbp)
L57:
  addl   $0x1,-0x74(%rbp)
L0:
  mov    -0x74(%rbp),%eax
  cmp    -0x64(%rbp),%eax
  jl     L27
L63:
  mov    -0x6c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
int Base64Decode(char *input, char *output, int oplen) {
  char *charval = 0;
  char decoderinput[4 + 1] = "";
  char encodingtabe[63 + 1] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
                              "abcdefghijklmnopqrstuvwxyz"
                              "0123456789"
                              "+/";
  ;
  int index = 0, asciival = 0, computeval = 0, iplen = 0, rc = 0;
  iplen = strlen(input);
  while (index < iplen) {
    asciival = (int)input[index];
    if (asciival == '=') {
      rc = decodeblock(decoderinput, output, oplen);
      break;
    } else {
      charval = strchr(encodingtabe, asciival);
      if (charval) {
        decoderinput[computeval] = charval - encodingtabe;
        computeval = (computeval + 1) % 4;
        if (computeval == 0) {
          rc = decodeblock(decoderinput, output, oplen);
          decoderinput[0] = decoderinput[1] = decoderinput[2] = decoderinput[3] = 0;
        }
      }
    }
    index++;
  }
  return rc;
}
```