C# 首次拟合算法输出

Question

我被分配了一个家庭作业来编写 first-fit 算法，它的意思是根据 first-fit 算法将一定数量的作业分配给一定数量的内存块。到目前为止，我的算法可以正常工作并正确分配工作；但是，我的问题是，当我排队的作业输出到控制台时，我的循环不会遍历所有发送到队列的作业，它只显示发送到队列的初始作业及其大小，直到所有要发送的作业对队列进行了迭代。

这可能非常简单，但我似乎无法弄明白...。任何其他需要的都可以提出，我可以提供。谢谢。

public void firstFit(int counter, int JobIndex)
  {

 Counter = counter;
 jobIndex = JobIndex;

do {
        if (jobSize[jobIndex] > memorySize[Counter-1])
    {  
        Counter += 1;
    }
    else
    { 
    Console.Text += Environment.NewLine + "Job " + (jobIndex+1) + " of size " 
    + jobSize[jobIndex] + 
    " has been loaded into memory block:"
    + Counter;

    memorySize[Counter-1] = memorySize[Counter-1]-jobSize[jobIndex];

    Console.Text += Environment.NewLine + "The size of memory block " + Counter + " is now " + memorySize[Counter-1];

    Counter = 1;
    jobIndex += 1;
    }

}   while (Counter <= blocks && jobIndex < jobs);

for (int i = 0; i < jobs-jobIndex; i++ ){

jobQ = jobIndex;

if (jobQ < jobs)
{
    Console.Text += Environment.NewLine + "Job " + (jobQ+1) + " of size " + jobSize[jobQ] + " is sent to waiting queue!";


 }
}

Job 1 of size 45 has been loaded into memory block:1 The size of memory block 1 is now 16

Job 2 of size 1230 has been loaded into memory block:2 The size of memory block 2 is now 410

Job 3 of size 325 has been loaded into memory block:2 The size of memory block 2 is now 85

Job 4 of size 303 has been loaded into memory block:3 The size of memory block 3 is now 131

Job 5 of size 1178 has been loaded into memory block:5 The size of memory block 5 is now 393

Job 6 of size 1276 has been loaded into memory block:6 The size of memory block 6 is now 426

Job 7 of size 965 has been loaded into memory block:7 The size of memory block 7 is now 322

Job 8 of size 537 is sent to waiting queue!

Job 8 of size 537 is sent to waiting queue!

Job 8 of size 537 is sent to waiting queue!

Answer 1

最后有这段代码:

for (int i = 0; i < jobs-jobIndex; i++ )
{
    jobQ = jobIndex;
    if (jobQ < jobs)
    {
        Console.Text += Environment.NewLine + "Job " + (jobQ+1) + " of size " +
            jobSize[jobQ] + " is sent to waiting queue!";
    }
}

jobQ 的值在每次循环中都是相同的，所以当然每次都会打印相同的内容。有没有可能你的意思是循环中的第一行是:

jobQ = jobIndex + i;

??