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12位唯一ID-代码可靠性

uniqueidentifier unique c++

Kamran Khan · 技术社区 · 17 年前

我想要一个在一天(24小时)内保持唯一的数字。下面是我想出的代码;我想知道它的谬误/可能的风险;'我相信“这保证了一天内至少有一个12位数的唯一数字。

逻辑是获取查询性能计数器结果的前四个字节的当前日期/时间(hhmmsmmm)和concat。

__forceinline bool GetUniqueID(char caUID[MAX_STRING_LENGTH])
{
    //Logic: Add HHMMSSmmm with mid 3 bytes of performance counter.
    //Guarantees that in a single milli second band (0 to 999) the three bytes 
    //of performance counter would always be unique.
    //1. Get system time, and use
    bool bStatus = false;
    try
    {

        SYSTEMTIME localtime;
        GetLocalTime(&localtime);//Get local time, so that we may pull out HHMMSSmmm

        LARGE_INTEGER li;
        char cNT[MAX_STRING_LENGTH];//new time.
        memset(cNT, '\0', sizeof(cNT));
        try
        {
            //Try to get the performance counter,
            //if one is provided by the OEM.

            QueryPerformanceCounter(&li);//This function retrieves the current value of the 
                                         //high-resolution performance counter if one is provided by the OEM
                                         //We use the first four bytes only of it.
            sprintf(cNT, "%u", li.QuadPart);
        }
        catch(...)
        {
            //Not provided by OEM.
            //Lets go with the GetTickCounts();
            //ddHHMMSS + 4 bytes of dwTicks
            sprintf(cNT,"%04d", GetTickCount());
        }


        //Get the first four bytes.
        int iSkipTo     = 0;//This is incase we'd decide to pull out next four bytes, rather than first four bytes.
        int iGetChars   = 4;//Number of chars to get.
        char *pSub = (char*) malloc(iGetChars+1);//Clear memory
        strncpy(pSub, cNT + iSkipTo, iGetChars);//Get string
        pSub[iGetChars] = '\0'; //Mark end.

        //Prepare unique id
        sprintf(caUID, "%02d%02d%02d%3d%s", 
                                    localtime.wHour, 
                                    localtime.wMinute, 
                                    localtime.wSecond, 
                                    localtime.wMilliseconds, 
                                    pSub); //First four characters concat.

        bStatus = true;
    }
    catch(...)
    {
        //Couldnt prepare. There was some problem.
        bStatus = false;
    }

    return bStatus;
}

以下是我得到的输出:

唯一:[125907 462224] 唯一:[125907 462227] 唯一:[125907 462228] 唯一:[125907 462230] 唯一:[125907 462233] 唯一:[125907 462235] 唯一:[12590746237] 唯一:[125907 462238] 唯一:[125907 462241] 唯一:[125907 462245] 唯一:[12590746246] 唯一:[125907 462248] 唯一:[12590746249] 唯一:[125907 462252] 唯一:[125907 462254] 唯一:[125907 462255] 唯一:[125907 462256] 唯一:[125907 462257] 唯一:[125907 462258] 唯一:[125907 622261] 唯一:[125907 622263] 唯一:[125907 622264] 唯一:[125907 622268] 唯一:[125907 622269] 唯一:[125907 622270] 唯一:[125907 622271] 唯一:[125907 622273] 唯一:[125907 622275] 唯一:[125907 622278] 唯一:[125907 622281] 唯一:[125907 622282] 唯一:[125907 622283] 唯一:[125907 622285] 唯一:[125907 622286] 唯一:[125907 622288] 唯一:[125907 622290] 唯一:[125907 622291] 唯一:[125907 622293] 唯一:[125907 622297] 唯一:[125907 622298] 唯一:[125907 622299] 唯一:[125907 622300] 唯一:[125907 622302] 唯一:[125907 622305] 唯一:[125907 782308] 唯一:[125907 782310] 唯一:[125907 782312] 唯一:[125907 782314] 唯一:[125907 782316] 唯一:[125907 782318] 毫秒改变,125 唯一:[1259071402495] 唯一:[1259071402497] 唯一:[1259071402498] 唯一:[1259071402499] 唯一:[1259071402502] 唯一:[1259071402503] 唯一:[1259071402504]

现在我考虑将生成的ID保存在一个列表中,并将新生成的ID与列表中现有的ID进行比较。如果它已经存在于列表中,那么我当然可以跳过这个数字并生成另一个,但显然这个逻辑会失败。

非常感谢您的意见/建议/更新等。

6 回复 | 直到 16 年前

Aaron Digulla 17 年前

static int counter = 0;
static Time lastTime;

String getNextId() {
    Time now = System.getTime();
    if (lastTime == now)
        counter ++;
    else
        counter = 0;
    return now+counter;
}

这将保证即使我调用该方法的次数超过 getTime() 变化。

lc. 17 年前

出于好奇,您不使用guid有什么原因吗?

或者,既然您正在考虑将生成的ID保存在一个列表中进行比较,为什么不能创建一个生成器?由于您建议存储是一个选项,如果您存储上次使用的数字并在每次使用时递增它。。。如果您愿意,您甚至可以存储日期并每天重置计数器。

Michael 17 年前

我在快速循环中遇到过一些问题,它们没有改变,尽管这是常识,它们应该改变(你无法控制操作系统,所以你不能假设时间每x毫秒改变一次,等等)

格式为TTTTOOO,其中T为时间数字,o为4位偏移量。

Kamran Khan 17 年前

__forceinline bool GetUniqueIDEx(char caUID[MAX_STRING_LENGTH])
{
    //Logic: Add HHMMSSmmm with 3 bytes counter.
    //Guarantees a unique number for a calendar date, 
    //that in a single milli second band (0 to 999) the three bytes 
    //of counter would always be unique.
    //1. Get system time, and use

    bool bStatus = false;
    try
    {
        GetLocalTime(&localtime);//Get local time, so that we may pull out HHMMSSmmm

        char cNT[MAX_STRING_LENGTH];//new time.
        memset(cNT, '\0', sizeof(cNT));
        if(m_nCounter> MAX_COUNTER_LIMIT)
        {
            m_nCounter= 0;
        }

        sprintf(cNT, "%03d", ++m_nCounter);

        //Prepare unique id
        sprintf(caUID, "%02d%02d%02d%03d%s", 
                                            localtime.wHour, 
                                            localtime.wMinute, 
                                            localtime.wSecond, 
                                            localtime.wMilliseconds, 
                                            cNT); 

        bStatus = true;
    }
    catch(...)
    {
        //Couldnt prepare. There was some problem.
        bStatus = false;
    }

    return bStatus;
}

jmucchiello 17 年前

sprintf(idstr, "%05d%07d", secs_since_midnight, counter);

当然,当你真的想把计数器塞进几个可打印的字符(和一年中的某一天,而不是月/日,等等)时,我也坚信使用base-36(通过itoa)

Manikanthan Velayutham 16 年前

我不确定您的流程是否可以牺牲微小的性能问题并保持逻辑简单。

如果您在两次呼叫之间将进程休眠1毫秒,我们可以保证HHMMSMM格式本身的唯一号码。通过这种方式,您可以消除连接部分,也可以消除为了双重检查唯一性而必须维护的列表。