user: What would the C source look like for the assembly code provided here?

```asm
<_base64_decode_triple>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3,-0x28(%rbp)
  movl   $0x1,-0x24(%rbp)
  movl   $0x0,-0x30(%rbp)
  jmp    L42
L148:
  mov    -0x30(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <_base64_char_value@plt>
  mov    -0x30(%rbp),%edx
  movslq %edx,%rdx
  mov    %eax,-0x20(%rbp,%rdx,4)
  addl   $0x1,-0x30(%rbp)
L42:
  cmpl   $0x3,-0x30(%rbp)
  jle    L148
  movl   $0x3,-0x30(%rbp)
  jmp    L77
L106:
  mov    -0x30(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  test   %eax,%eax
  jns    L19
  cmpl   $0x0,-0x24(%rbp)
  je     L129
  mov    -0x30(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x3d,%al
  jne    L129
  mov    -0x30(%rbp),%eax
  cltq
  movl   $0x0,-0x20(%rbp,%rax,4)
  subl   $0x1,-0x28(%rbp)
  jmp    L68
L129:
  mov    $0x0,%eax
  jmp    L34
L19:
  movl   $0x0,-0x24(%rbp)
L68:
  subl   $0x1,-0x30(%rbp)
L77:
  cmpl   $0x0,-0x30(%rbp)
  jns    L106
  cmpl   $0x0,-0x28(%rbp)
  jns    L49
  movl   $0x0,-0x28(%rbp)
L49:
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x2c(%rbp)
  shll   $0x6,-0x2c(%rbp)
  mov    -0x1c(%rbp),%eax
  add    %eax,-0x2c(%rbp)
  shll   $0x6,-0x2c(%rbp)
  mov    -0x18(%rbp),%eax
  add    %eax,-0x2c(%rbp)
  shll   $0x6,-0x2c(%rbp)
  mov    -0x14(%rbp),%eax
  add    %eax,-0x2c(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,-0x30(%rbp)
  jmp    L104
L132:
  mov    -0x2c(%rbp),%eax
  lea    0xff(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x8,%eax
  mov    %eax,-0x2c(%rbp)
  addl   $0x1,-0x30(%rbp)
L104:
  cmpl   $0x2,-0x30(%rbp)
  jle    L132
  mov    -0x28(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x30(%rbp)
  jmp    L23
L31:
  mov    -0x2c(%rbp),%eax
  cltd
  shr    $0x18,%edx
  add    %edx,%eax
  movzbl %al,%eax
  sub    %edx,%eax
  mov    %eax,%ecx
  mov    -0x30(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x2c(%rbp),%eax
  lea    0xff(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x8,%eax
  mov    %eax,-0x2c(%rbp)
  subl   $0x1,-0x30(%rbp)
L23:
  cmpl   $0x0,-0x30(%rbp)
  jns    L31
  mov    -0x28(%rbp),%eax
L34:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L50
  call   <__stack_chk_fail@plt>
L50:
  leave
  ret
```

assistant: ```c
int _base64_decode_triple(char quadruple[4], unsigned char *result) {
  int i, triple_value, bytes_to_decode = 3, only_equals_yet = 1;
  int char_value[4];
  for (i = 0; i < 4; i++)
    char_value[i] = _base64_char_value(quadruple[i]);
  for (i = 3; i >= 0; i--) {
    if (char_value[i] < 0) {
      if (only_equals_yet && quadruple[i] == '=') {
        char_value[i] = 0;
        bytes_to_decode--;
        continue;
      }
      return 0;
    }
    only_equals_yet = 0;
  }
  if (bytes_to_decode < 0)
    bytes_to_decode = 0;
  triple_value = char_value[0];
  triple_value *= 64;
  triple_value += char_value[1];
  triple_value *= 64;
  triple_value += char_value[2];
  triple_value *= 64;
  triple_value += char_value[3];
  for (i = bytes_to_decode; i < 3; i++)
    triple_value /= 256;
  for (i = bytes_to_decode - 1; i >= 0; i--) {
    result[i] = triple_value % 256;
    triple_value /= 256;
  }
  return bytes_to_decode;
}
```