Question

使用Bitwise Or（包含或）和Shift左[＆lt;＆lt;＆lt;＆lt;]运算符在PHP中编码的时间。如何解码结果以获取Back Year，TotalDays，totalSeconds？

$Year=17;   // means 2017
$TotalDays=223;
$totalSeconds=5435;

$Result = ( (intval($Year) << 26) | (intval($TotalDays) << 17) | intval($totalSeconds));

编码hermes时间的orignal算法如下编码hermes时间的orignal算法如下所示

unsigned int Gps_DateTimeEncode(short Hours,short Minutes,short Seconds,short Year,short Month, short Days)
{
  unsigned int Result;
  int Months[12]= {31,28,31,30,31,30,31,31,30,31,30,31};
  int TotalSeconds = (Hours *3600) + (Minutes * 60) + Seconds;
  int TotalDays=0,i;
  if(Year%4==0)  Months[1]=29;

  for(i=0;i<Month-1;i++)
  {
    TotalDays+=Months[i];
  }
  TotalDays+=Days;
  Result =  ((Year) << 26) | (TotalDays << 17) | (TotalSeconds) ;
  return Result;                                                    
}

Answer 1

/**
 * Takes an encoded date and returns a \stdClass object with 'year', 'days'
 * and 'seconds' properties decoded
 * @param int $encoded Encoded date
 * @return \stdClass
 */
function decodeHermesTime($encoded){

    $obj = new \stdClass();

    // right-shift 26 spots to get raw year
    $obj->year = $encoded >> 26;

    // subtract year bits from $encoded to get daysAndSecondsBits
    $daysAndSecondsBits = $encoded - ($obj->year << 26);

    // right-shift 17 times to get days
    $obj->days = $daysAndSecondsBits >> 17;

    // subtract days bits from $daysAndSecondsBits to have raw seconds
    $obj->seconds = $daysAndSecondsBits - ($obj->days << 17);

    return $obj;
}

// usage:
$Result = 17 << 26 | 223 << 17 | 5435;
$r = decodeHermesTime($Result);
echo "Year: $r->year, Days: $r->days, Seconds: $r->seconds";
// Year: 17, Days: 223, Seconds: 5435

Live Demo

使用Bitwise运算符在PHP中编码的时间。如何解码结果？

1 个答案: