Swapping by reference array values

Question

So right after I do this swap function all the values in my array goes to -858993460. Why that number? I think it's the way I declare my array, but isn't A a pointer by declaring it that way? I'm trying to figure out why this is. well here's the code:

int* arrayCreate()//added mg
{
    int a[] = { 3, 7, 4, 9, 5, 2 };

    return a;
}

void Sort::Insert()
{   
    int t = 0;

    //start clock
    clock_t start = clock();
    d = arrayCreate();  
    size = 6;
    for (int i = 1; i< size; i++)
    {
        t = i;

        while (((t - 1) >= 0) && (d[t] < d[t- 1]))
        {
            //Swap
            Swap(d[t], d[t- 1]);  // right after this call 
                                  //the debugger says values go to - 8e^8
            t--;
        }
    }
}

void Sort::Swap(int& n1, int& n2)
{
    int temp = n1;
    n1 = n2;
    n2 = temp;
}

Doing some testing I realize if I do: int *a = new int[6]; a[0] = 3; a[1] = 7; a[2] = 4; a[3] = 9; a[4] = 5; a[5] = 2; This doesn't screw up the array after swap. But why? I am so curious I really must know. — zissler
– zissler, Commented Feb 21, 2015 at 6:13

Freddy · Accepted Answer · 2015-02-21 06:15:02Z

2

You're overlooking that all stack based variables are destroyed once the scope they are defined in is exited.

therefore, as soon as you return A, it points to memory that is undefined.

int* arrayCreate()//added mg
{
    int a[] = { 3, 7, 4, 9, 5, 2 };

    return a;
}

Could be changed to this

int* arrayCreate()//added mg
{
    static int a[] = { 3, 7, 4, 9, 5, 2 };

    return a;
}

Marking the array as static will cause the memory to stick around.

answered Feb 21, 2015 at 6:15

Freddy

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4 Comments

zissler Over a year ago

It makes a lot of sense, but the values do stick around until swap in the debugger. Why is that?

Freddy Over a year ago

Undefined is undefined. If you compile your code with a different compiler you will most likely get different results. Please know that when I say undefined I am not talking about the memory itself, but what the language says that data should be. It is implementation and compile dependent and the results will very. In CPP this is commonly referred to as undefined behavior.

zissler Over a year ago

I guess visual studios was throwing me off. I'll go with what you said. Also then is there an easier way to declare a predefined array pointer without using static?

Freddy Over a year ago

You could always make it global (outside any functions).

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