Question

我正在尝试编写一个模拟LRU页面替换的函数。我很了解LRU，但编码时遇到了问题。以下内容正在传递到LRU函数中。用户指定＃1-9的20字符引用字符串，该字符串存储在名为refString的大小为20的数组中。用户输入的帧数（1-7）存储在变量numFrames中。最后，传入一个大小为7的数组，称为帧。

这是我的代码，我得到一个接近的数字但不完全。也许有人可以帮忙！

private static void LRU(int numFrames, int[] refString, int[] frame)
{
    int i, j = 0, k, m, flag = 0, count = 0, top = 0;

    for (i = 0; i < 20; i++)
    {
        for (k = 0; k < numFrames; k++)
        {
            if (frame[k] == refString[i])
            {
                flag = 1;
                break;
            }
        }

        if (j != numFrames && flag != 1)
        {
            frame[top] = refString[i];
            j++;

            if (j != numFrames)
            {
                top++;
            }
        }

        else
        {
            if (flag != 1)
            {
                for (k = 0; k < top; k++)
                {
                    frame[k] = frame[k + 1];
                }

                frame[top] = refString[i];
            }

            if (flag == 1)
            {
                for (m = k; m < top; m++)
                {
                    frame[m] = frame[m + 1];
                }

                frame[top] = refString[i];
            }
        }

        if (flag == 0)
        {
            count++;
        }
        else
        {
            flag = 0;
        }

    }

    Console.WriteLine("\nThe number of page faults with LRU is: " + count);
}

Answer 1

您的代码中存在少量错误： -

if (top < numFrames)
        {
            frame[top++] = refString[i];
            fault++;
        }

这里你永远不会检查当前的refString [i]是否已经在frame []中，因为在这种情况下你不会出错并且不应该在帧中添加它。

这是一个伪代码，可以帮助您清除疑虑： -

void LRU(int numframes,int refString[],int frames[]) {

   int top = 0,fault=0;
   int* count = new int[numframes];

   for(int i=0;i<refString.length;i++) {

       int k = findmax(refString[i],frames,count,top,numframes);

       if(k<0) {
          count[top] = 0;
          frames[top++] = refString[i];
          fault++;   
       }

       else if(frames[k]!=refString[i]) {

           count[k] = 0;
           frames[k] = refString[i];
           fault++;

       }
      else count[k] = 0;

     for(int j=0;j<top;j++) {
          count[j]++;  

     }

   }

   return(fault);

}


int findmax(int keyframe,int frames[],int count,int top,int numframes) {

     int max = 0;
     for(int i=0;i<top;i++) {

        if(frames[i]==keyframe) {

             return(i);
        }
        if(count[max]<count[i])
            max = i;

     }

     if(top<numframes)
           return(-1);
     return(max);
}

修改

伪代码说明： -

1. check if current requested frame is in cache and if yes then get its index 2. if frame is present then set its count to zero so as to indicate it is used very recently, higher the count the more least recently frame is used. 3. if frame is not present in cache then a. check if cache is full if not add new frame to end of cache and increment the fault and top of cache b. else if chace is full then get frame with maximum count(LRU frame) and replace it with new frame and reset count to zero and increment fault. 4. increment all the counts 5. do 1 to 4 till end of all requests 6. output the no of faults

LRU页面替换算法C＃

1 个答案: