The "bubble sort algorithm" small error

Question

I was given an assignment to:

"Write a function that implements the bubble sort for an array of integers. The prototype for the function is as follows: void sort(int data[], int count);

Write a program that prompts for a file name, opens that file and reads numbers from the file in to an array. Your program should then call your sort routine and then print the resulting array."

I completed it, and it was working just fine until I was told it also has to stop early if the numbers are already in order. You will see that attempt in the void function below. I'm not getting any errors; the program compiles, executes, and closes fine. The problem is now with the addition of the boolean variable it no longer sorts the numbers. It will print them out in the same order as the text file. I ran it through the debugger and noticed that the "ii" variable wasn't increasing by 1 every pass like it's supposed to , for(... ; ... ; ii++) , but instead stayed at 0. Any ideas?

The (random) numbers inside the "integers.txt" file: 12 42 5 67 41 9 19 93 10 124 21

void sort(int data[], int count);

int main()
{
    const int MAX_SIZE = 128;
    char fileName[MAX_SIZE];
    int data[MAX_SIZE];
    int count = 0;

    cout << "Enter a file name: "; //integers.txt
    cin.clear();
    cin.ignore(cin.rdbuf()->in_avail());
    cin.getline(fileName, sizeof(fileName));

    ifstream input(fileName); //opens the file and reads the first line
    if (input.is_open())
    {
        while (!input.eof() && count <= MAX_SIZE) //adds data to the array until the end of the file is reached
        {
            input >> data[count];
            count += 1;
        }
    input.close();
    }
    else
    {
        cout << "\nThe file failed to open, try again.\n";
    }

    sort(data, count); //calls the bubble sort function

    return 0;
}

void sort(int data[], int count)
{
    int temp = 0;
    int pass = 0;
    bool sorted = false;

    for (int pass = 0; pass <= count; pass += 1) //counts the number of passes
    {
        for (int ii = 0; (ii <= (count - pass - 1)) && (sorted = false) ; ii++) //sorts the integers from least to greatest
        {                                                                       //also 'supposed to' stop early if already sorted
            if (data[ii] > data[ii + 1])
            {
                sorted = false;
                temp = data[ii];
                data[ii] = data[ii + 1];
                data[ii + 1] = temp;
            }
        } 
    }

    cout << "\nSorted integers: ";

    for (int jj = 1; jj <= count; jj += 1) //prints the sorted integers
    {
        cout << data[jj] << " ";
    }

    cout << "\n\n";
}

i suspect `sorted = false` should be `sorted == false` or even more idiomatically `!sorted` — pm100, Nov 20 '17 at 21:34
Please read this https://stackoverflow.com/questions/5605125/why-is-iostreameof-inside-a-loop-condition-considered-wrong — Ed Heal, Nov 20 '17 at 21:35

score 0 · Accepted Answer · answered Nov 20 '17 at 21:38

You want to make the sorted check in the outer loop. That is:

for (int pass = 0; pass < count && !sorted; pass += 1) //counts the number of passes
{
    sorted = true;
    for (int ii = 0; ii < (count - pass); ii++) //sorts the integers from least to greatest
    {

The idea here is that at the beginning of each pass, you assume that the array is sorted. If you make an exchange, then the array is potentially not sorted, so you set the flag to false. The flag will only be true if you make an entire pass without any exchanges.

Also, note that I changed your outer loop comparison to pass < count rather than pass <= count. Remember, when you're starting from 0, the limit is count-1.

I also changed the inner loop condition from <= (count - pass - 1) to < (count - pass). They're equivalent, but the latter is more concise.

The "bubble sort algorithm" small error

1 Answers1