Question

我担心的是，在线程之间访问全局指针会对全局指针产生什么影响。我的全局指针是线程安全类。当updatethread()方法使用新指针更新指针并且workerthread()访问指针时，从代码中将对全局指针产生什么影响。我应该使用哪种同步方式？

SomeCache* ptrCache = NULL;

    //worker thread
    void Workerthread(std::string strFileToWork)
    {
        while (std::getline(fileStream, strCurrent))
        {                   
            //worker thread accessing the global pointer
        if (ptrCache->SearchValue(strCurrent))
        {           
            iCounter++;
        }
        }
    }

    void updatethread()
    {
        //replace old cache with a new fresh updated.
        SomeCache* ptrOldCache = ptrCache;
        ptrCache = ptrNewCache;
    }

Answer 1

使用App es的一种可能的解决方案：

var a = [1, 4, 9, 78, 42, 4, 11, 4, 9];

result = [];
verified = [];
for(var i =0; i<a.length; i++){
   b = a[i];
   temp = {};
   temp.value = a[i];
   temp.hasDuplicate = false;
   temp.positions = [];
   temp.differenceFromFirstOccurence = [];
   if(verified.indexOf(b) === -1){
      temp.positions = [i+1];
      for(var j = 0; j <a.length; j++){
        c = a[j];
            if( i !== j && b === c){
         temp.hasDuplicate = true;
         temp.positions.push(j+1);
        }
        }
      verified.push(b);
      result.push(temp);

   }
}

for(var i = 0; i < result.length; i++){
  if(result[i].hasDuplicate){
    firstPosition = result[i]['positions'][0];
    for(var j = result[i]['positions'].length-1; j > 0; j--){
       diff = result[i]['positions'][j] - firstPosition-1;
        result[i].differenceFromFirstOccurence.push(diff);
    }
  }
  result[i].differenceFromFirstOccurence.reverse();
}

console.log(result);

for(var i =0; i < result.length; i++){
    if(result[i].hasDuplicate && result[i].differenceFromFirstOccurence.length){

            console.log("The first occurence of "+result[i].value+" is at "+ result[i].positions[0]);
      for(var j = 1; j < result[i].positions.length; j++){
       console.log("The duplicate occurence of "+result[i].value+" is at "+ result[i].positions[j] +" and difference is "+ result[i].differenceFromFirstOccurence[j-1]);
      }
    }
}

如果编译器不支持模板参数推导，则应明确指定互斥锁类型：Select DateSerial(TYear ,TMonth, TDay) As TDate From NDX_FULL。

Answer 2

行为可能是不确定的，您可以参考有关C ++中 volatile 关键字的答案。 https://stackoverflow.com/a/72617/10443813 如果使用 volatile 关键字，则在执行以下行之前，将使用旧值，而在之后，将使用新值。否则，行为可能取决于编译器或编译标志。

    ptrCache = ptrNewCache;

线程更新时对全局指针的影响

2 个答案: