arranging array in ascending order

Question

I'm trying to arrange a random array. In the code it's just the first step of swapping places by size. When running the code a get Debug Error and the output after the swap shows that the first number in the array was deleted and the last is a long random number that was in the memory. It looks like it starts the swapping from i=1, why?

#include <iostream>
#include <cstdlib>
#include <ctime>

using namespace std;

void buildArray(int arr[], int size) {
    srand ( (unsigned int)time(NULL) );
    for(int i = 0; i < size; ++i)
        arr[i] = rand() % 50 + 0;
}


void dispArray(int arr[], int size) {
     for(int i = 0; i < size; ++i)
        cout << i << ": " << arr[i] << endl;
}


int main()
{
        const int size = 5;
    int     arr[size];
    buildArray(arr, size);
    dispArray(arr, size);

    int swapHolder = -1;
    for(int i = 0; i < size; ++i) {
    if(arr[i] > arr[i+1]) {
            swapHolder = arr[i+1];
            arr[i+1] = arr[i];
            arr[i] = swapHolder;
            cout << endl;
        }
    }
    dispArray(arr, size);

    return 0;
}  


Output example:
0: 46
1: 15
2: 47
3: 5
4: 19


0: 15
1: 46
2: 5
3: 19
4: -858993460

yizzlez · Accepted Answer · 2014-03-13 01:00:38Z

2

Here is the problem: if(arr[i] > arr[i+1]), your for loop goes from [0, size - 1], so on the last iteration, lets say size = 5, you would be testing if(arr[5] > arr[6]), accessing the uninitialized arr[6], the right way is to make your for loop go from [0, size - 2]:

for(int i = 0; i < size - 1; ++i) ...

answered Mar 13, 2014 at 1:00

yizzlez

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4pie0 · Accepted Answer · 2014-03-13 01:02:37Z

1

for(int i = 0; i < size; ++i) {
    if(arr[i] > arr[i+1]) {
            swapHolder = arr[i+1];
            arr[i+1] = arr[i];

The last iteration is for i = size -1. Then

 if(arr[i] > arr[i+1])

means accessing array out of range. This is undefined behavior.

answered Mar 13, 2014 at 1:02

4pie0

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