5

我目前正在为 natvis 中的日期类型编写可视化工具。date 类型以通常的 unix 方式存储自 1970 年以来的秒数，但如果不使用临时变量，则从中派生年、月和月中的日期会非常冗长。我希望能够存储工作变量，以便以理智的方式评估可视化器。这可能吗？

我到达的解决方案如下所示：

struct SLowResAbsoluteTime
{
    SLowResAbsoluteTime() : mTime(0) { calcDOE(); }
    SLowResAbsoluteTime(int year, SDate::EMonth m, SDate::EDayOfWeek day, UINT8 hour, UINT8 minute, UINT8 seconds);
    SLowResAbsoluteTime(const SDate &date);
    SLowResAbsoluteTime(unsigned long long time) : mTime(time) { calcDOE(); }
    SLowResAbsoluteTime(const SLowResAbsoluteTime &other) : mTime(other.mTime) { calcDOE(); }

    SDate getDate() const; //calculate date object from given time
    UINT32 getHour() const; 
    UINT32 getMinutes() const;
    UINT32 getSeconds() const;

    bool operator < (const SLowResAbsoluteTime &other) const { return mTime < other.mTime; }
    bool operator > (const SLowResAbsoluteTime &other) const { return mTime > other.mTime; }
    bool operator <= (const SLowResAbsoluteTime &other) const { return mTime <= other.mTime; }
    bool operator >= (const SLowResAbsoluteTime &other) const { return mTime >= other.mTime; }
    bool operator == (const SLowResAbsoluteTime &other) const { return mTime == other.mTime; }
    bool operator != (const SLowResAbsoluteTime &other) const { return mTime != other.mTime; }
    SLowResAbsoluteTime operator -(const SLowResAbsoluteTime &time) const { return SLowResAbsoluteTime(mTime - time.mTime); }
    SLowResAbsoluteTime operator +(const SLowResAbsoluteTime &time) const { return SLowResAbsoluteTime(mTime + time.mTime); }
    const SLowResAbsoluteTime &operator -=(const SLowResAbsoluteTime &time) { mTime -= time.mTime; return *this; }
    const SLowResAbsoluteTime &operator +=(const SLowResAbsoluteTime &time) { mTime += time.mTime; return *this; }
    unsigned long long mTime;
    void invalidate() { mTime = -1; }
    bool isValid() const {return mTime != UINT64(-1); }
    operator unsigned long() const { return (long)mTime; }
    void calcDOE();
#ifdef USING_DEBUG_TIMER_DOE
    struct { UINT16 y; UINT8 m; UINT8 d; } mDOE;
#endif
};

请注意“USING_DEBUG_TIMER_DOE”部分。是这样计算的：

void SLowResAbsoluteTime::calcDOE()
{
#ifdef USING_DEBUG_TIMER_DOE
    int ts = mTime / (60 * 60 * 24);
    int z = ts + 719468;
    int doe = (z - ((z >= 0 ? z : z - 146096) / 146097) * 146097);
    int yoe = (doe - doe / 1460 + doe / 36524 - doe / 146096) / 365;  // [0, 399]
    int era = (z >= 0 ? z : z - 146096) / 146097;
    int y = (yoe) + era * 400;
    int doy = doe - (365 * yoe + yoe / 4 - yoe / 100);                // [0, 365]
    int mp = (5 * doy + 2) / 153;                                   // [0, 11]
    int d = doy - (153 * mp + 2) / 5 + 1;                             // [1, 31]
    int m = mp + (mp < 10 ? 3 : -9);                            // [1, 12]
    mDOE.y = y + (m <= 2);
    mDOE.m = m;
    mDOE.d = d; 
#endif  
}

用于可视化这些的 natvis 是：

<Type Name="SLowResAbsoluteTime">

    <DisplayString>{{time = { (mTime / (60 * 60)) % 24  }:{(mTime / 60) % 60}:{mTime % 60} } day-1970: {mTime / (60 * 60 * 24)} }</DisplayString>

    <Expand>        
        <Item Name="month">(int)mDOE.m</Item>
        <Item Name="day">(int)mDOE.d</Item>-->
        <Item Name="secs since 1/1/1970"> mTime</Item>  
    </Expand>
</Type>
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1 回答 1

5

如果您想从您的代码中消除calcDOE和mDOE并在适当的 natvis 中进行计算 - 是的，这是可能的。使用Intrinsic功能：

  <Type Name="SLowResAbsoluteTime">
    <Intrinsic Name="ts"   Expression="mTime / (60 * 60 * 24)                                       "/>
    <Intrinsic Name="z"    Expression="ts() + 719468                                                "/>
    <Intrinsic Name="doe"  Expression="(z() - ((z() &gt;= 0 ? z() : z() - 146096) / 146097) * 146097)"/>
    <Intrinsic Name="yoe"  Expression="(doe() - doe() / 1460 + doe() / 36524 - doe() / 146096) / 365"/>
    <Intrinsic Name="era"  Expression="(z() &gt;= 0 ? z() : z() - 146096) / 146097                  "/>
    <Intrinsic Name="y"    Expression="yoe() + era() * 400                                          "/>
    <Intrinsic Name="doy"  Expression="doe() - (365 * yoe() + yoe() / 4 - yoe() / 100)              "/>
    <Intrinsic Name="mp"   Expression="(5 * doy() + 2) / 153                                        "/>
    <Intrinsic Name="d"    Expression="doy() - (153 * mp() + 2) / 5 + 1                             "/>
    <Intrinsic Name="m"    Expression="mp() + (mp() &lt; 10 ? 3 : -9)                               "/>
    <Intrinsic Name="DOEy" Expression="y() + (m() &lt;= 2)                                          "/>
    <Expand>
      <Item Name="day">d()</Item>
      <Item Name="month">m()</Item>
      <Item Name="year">DOEy()</Item>
    </Expand>
  </Type>
于 2020-05-07T10:58:20.720 回答