C ++顺序搜索未找到最后一个元素

Question

So I have an input file. It consists of 40 numbers. The first 20 numbers are input into an array (I've checked this, they're actually there). I then close and re-open the input file. I compare the first 20 numbers in the input file against my array using sequential search. This means they should all be successful. I then compare the next 20 numbers against the numbers in my array, they should all be unsuccessful searches. My array is unsorted at this point.

The problem I'm running into is that the last number for successful is never found using sequential. I'm not sure how to fix this.

Here is the sequential search function:

length = 19;

void Search::sequential(ItemType item, bool& found)
{ 
  int place = 0;
  while (place < length && item != list[place])
    place++;
  found = (place < length); 
}

And here is my successful/unsuccessful loops

outFile << "\n\n ************Sequential Successful ********** \n";
outFile << endl << "ID" << endl;

inFile >> num;
for(int i=0; i<=length && inFile; i++)
{
  search.sequential(num, found);
  if (found)
    outFile << num << endl; 

  inFile >> num;
} 


//sequential unsuccessful
outFile << "\n\n ************Sequential unsuccessful ********** \n";
outFile << endl << "ID" << endl;

for(int i=0; i<=length && inFile; i++)
{
  search.sequential(num, found);
  if (!found)
    outFile << num << endl; 

  inFile >> num;
}

However, my output is:

 ************Sequential Successful ********** 

 ID
 1111
 3352
 4567
 5678
 6789
 7890
 8901
 9012
 1223
 2113
 8546
 2374
 4723
 9573
 3284
 7474
 8594
 3589
 5858
 //THERE SHOULD BE 1925 HERE BUT THERE ISN'T

  ************Sequential unsuccessful ********** 

 ID
 9456
 3584
 2222
 4319
 4477
 5710
 5497
 1502
 1599
 1504
 1506
 9943
 8833
 9944
 6678
 5555
 5660
 9911
 6130
 1613

If I remove the "if (found)" statement everything works perfectly, but how do I get around this without removing that?

Thanks in advance

---------------edit---------------

Okay, when I changed length to 20 it still didn't seem to work. I'm so lost.

Here is where I create the array

inFile >> num;
for (int i=0; i<length && inFile; i++)
{
  search.addToList(num);
  inFile >> num;
}

and here is the addToList function

 void Search::addToList(ItemType num)
 {
   if (index < length)  //ive tried taking out this statement just to see if it makes a difference and it didn't
   {
     list[index] = num;
     index++;
   }
 }

I initialize index to 0 in the constructor

This is how I declare the array

    ItemType list[length]; 

IT WORKS!!!! Thank you all SO much! I really appreciate it so much.

Answer 1

There are 2 solutions : length should get 20 as value

length = 20;

or use "<=" instead of "<" (in this case "length" should be named "lastIndex")

void Search::sequential(ItemType item, bool& found) 
{  
  int index = 0; 
  while (index <= length && item != list[index]) 
    index++; 
  found = (index <= length);  
} 

Answer 2

INDEX SEQUENTIAL SEARCH USING C

this code works for all the cases ie if we are finding last element in an array this code will work...

#include<stdio.h>
void main()
{
  int d[100],kin[20],pin[20],temp,k,i,j=0,n,n1=0,start,end;
  printf("Enter the number of elements:");
  scanf("%d",&n);
  for(i=0;i<n;i++)
    scanf("%d",&d[i]);
  printf("Enter the number to be searched:");
  scanf("%d",&k);
  for(i=0;i<n;i+=3)
  {
    kin[n1]=d[i];
    pin[n1]=i;
    n1++;
  }
  if(k < kin[0])
  {
    printf("element not found");
    exit(0);
  }
  else
  {
    for(i=1;i<=n1;i++)
      if(k < kin[i] )
      {
        start=pin[i-1];
        end=pin[i];
        break;
      }
      else
      {
        start=n1;
        end=n-1;
      }
  }
  for(i=start;i<=end;i++)
  {
    if(k==d[i])
    {
      j=1;
      break;
    }
  }
  if(j==1)
    printf("element found at position %d",i);
  else
    printf("element not found");
}

Answer 3

看一下您的搜索功能，当您尝试找到第20个数字时，索引将具有什么值？

Answer 4

If you've got 20 numbers, then why do you set length to 19? That's very counter-intuitive.

Answer 5

Classic Off-By-One issue. See @Kipotlov's answer for code corrections.