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I read the dynamic array and use it, but i have a bit question: When i put following format in prototype:

void mmyfunc(int *& myArray)
{
 //implementation
}

1.if i pass an array to it, how to call it? because i get :

no matching function for call to

2.When i use the following implementation : 

void NetworkSocket::resizeArray (int *&orig, int index, int size)
{
    int *resized =  new  int[size];
    for (int i = 0; i < size; i++)
    {
        if ( i == index )
            i++;

        resized[i] = orig[i];
    }
    delete [] orig;
    orig = (int *)new int[size];
    orig = resized;
}

i get seg fault in delete [] line.

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2
  • 10
    How are you passing an "array" to it? Also, this code is leaks and segfaults all over: why go out of your way to get extra trouble? Use std::vector. Commented Jul 30, 2012 at 17:34
  • 1
    You left out the most important part of the code ... how you call it. Commented Jul 30, 2012 at 17:37

1 Answer 1

Reset to default
1

I'll just assume by array you mean

int x[16];

and you call the function as 

resizeArray (x, ...)

So delete[]-ing it explicitly results in undefined behavior, because it wasn't allocated with new[]. Moreover, x resides in automatic memory, so it gets cleaned up by itself. Moreover moreover, you can't re-assign arrays. Arrays are not pointers. They decay to pointers when passed as arguments.

If you declared the array as 

int* x = new int[16];

then your approach would work. But not quite C++ yet.

You can just use a std::vector and resize(). Don't re-invent the wheel (unless this is an assignment).

EDIT: Just spotted this:

orig = (int *)new int[size];
orig = resized;

will leak. Remove the extra new[].

EDIT 2:

What's 

if ( i == index )
        i++;

supposed to do?

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3 Comments

I would have put the vector advice first, but that's just me.
@EtiennedeMartel could be for educational purposes.
"then your approach would work"... The general approach yes, but not the current implementation.

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