user: 请从给定的汇编指令推导出C源代码:

```asm
<Base64Decode>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  movdqa D17(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x1b(%rsp)
  movaps %xmm0,0x20(%rsp)
  movdqa D14(%rip),%xmm0
  movb   $0x0,0x1f(%rsp)
  movaps %xmm0,0x30(%rsp)
  movdqa D5(%rip),%xmm0
  movaps %xmm0,0x40(%rsp)
  movdqa D37(%rip),%xmm0
  movaps %xmm0,0x50(%rsp)
  call   <strlen@plt>
  test   %eax,%eax
  jle    L51
  sub    $0x1,%eax
  movl   $0x0,0xc(%rsp)
  xor    %ebx,%ebx
  lea    0x20(%rsp),%r12
  lea    0x1(%r15,%rax,1),%rbp
  jmp    L42
  nopw   0x0(%rax,%rax,1)
L54:
  add    $0x1,%r15
  cmp    %rbp,%r15
  je     L9
L42:
  movsbl (%r15),%esi
  cmp    $0x24,%esi
  je     L18
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L54
  movslq %ebx,%rdx
  sub    %r12,%rax
  add    $0x1,%ebx
  mov    %al,0x1b(%rsp,%rdx,1)
  and    $0x3,%ebx
  jne    L54
  lea    0x1b(%rsp),%rdi
  mov    %r14d,%edx
  mov    %r13,%rsi
  add    $0x1,%r15
  call   <decodeblock@plt>
  movl   $0x0,0x1b(%rsp)
  mov    %eax,0xc(%rsp)
  cmp    %rbp,%r15
  jne    L42
  xchg   %ax,%ax
L9:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L83
  mov    0xc(%rsp),%eax
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  lea    0x1b(%rsp),%rdi
  mov    %r14d,%edx
  mov    %r13,%rsi
  call   <decodeblock@plt>
  mov    %eax,0xc(%rsp)
  jmp    L9
L51:
  movl   $0x0,0xc(%rsp)
  jmp    L9
L83:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D17
data_type: byte[16]
name: retrieve_data
arguments: tag: D14
data_type: byte[16]
name: retrieve_data
arguments: tag: D5
data_type: byte[16]
name: retrieve_data
arguments: tag: D37
data_type: byte[16]

tool: {"D17": [0x41,0x42,0x43,0x44,0x45,0x46,0x47,0x48,0x49,0x4a,0x4b,0x4c,0x4d,0x4e,0x4f,0x50]}

tool: {"D14": [0x51,0x52,0x53,0x54,0x55,0x56,0x57,0x58,0x59,0x5a,0x61,0x62,0x63,0x64,0x65,0x66]}

tool: {"D5": [0x67,0x68,0x69,0x6a,0x6b,0x6c,0x6d,0x6e,0x6f,0x70,0x71,0x72,0x73,0x74,0x75,0x76]}

tool: {"D37": [0x77,0x78,0x79,0x7a,0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x00,0x00]}

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);
}
```