Writing c++ "array" class myself

Question

I am writing c++ integer array class myself and a function for extracting sub-arrays but I am getting this error while trying to compile my code:

"Multiple markers at this line - No return, in function returning non-void - with ‘Array Array::subArr(int, int)’"

(sorry forgot to include this error)

- ‘int* Array::subArr(int, int)’ cannot be overloaded

But why cant it be overloaded?

The private data members of class Array are:

private:
    int *arr;
    int *subArray;
    int *subArr1;
    int len;

The two public functions causing the error are:

int * subArr(int pos, int siz)
{
    if (pos+siz < len)
    {
        subArr1 = new int[siz];
        for (int i=0; i<siz; i++)
        {
            subArr1[i] = arr[pos];
            pos++;
        }
        return subArr1;
    }
}

and

Array subArr(int pos, int siz)
{
    if (arr != NULL && (siz + pos) < len)
    {
        Array sub;
        sub.len = siz;
        sub.subArray = new int [siz];
        for (int i=0; i<siz; i++)
        {
            sub.subArray[i] = arr[pos];
            pos++;
        }
        return sub;
    }
    cout<<"Error. Enter a valid index and/or position\n";
}

I can send the whole class code if it is required.

Well, what's unclear about the error message? Outside the `if` block there's no `return`. — πάντα ῥεῖ, Mar 01 '16 at 16:24
the error message is actually quite clear, your function is supposed to return an `int *`, but for some cases it does not. — 463035818_is_not_an_ai, Mar 01 '16 at 16:24
Did you read what the Compiler says? It says your function does not return a value. And clearly there are code pathes (`else` because your `if` is not fulfilled) where the function does not have a return. — Werner Henze, Mar 01 '16 at 16:24

DevSolar · Accepted Answer · 2016-03-01T16:44:34.953

It's just what it says;

No return, in function returning non-void - with ‘Array Array::subArr(int, int)’"

Look at the code:

int * subArr(int pos, int siz)
{
    if (pos+siz < len)
    {
        subArr1 = new int[siz];
        for (int i=0; i<siz; i++)
        {
            subArr1[i] = arr[pos];
            pos++;
        }
        return subArr1;
    }
}

What happens if the if (pos+siz < len) is false?

Correct, you don't have anything to return, in a function declared to return int *. That's an error.

‘int* Array::subArr(int, int)’ cannot be overloaded

Look at the code:

int * subArr(int pos, int siz)
Array subArr(int pos, int siz)

The two functions differ only in return type. That is not possible; there must be a difference in the number and / or the type of the parameters.

Gratuitous generic advice:

Avoid reimplementing standard functionality, other than as exercise.
Avoid using pointers.
Avoid new unless...
- in a constructor where your destructor has the corresponding delete.
- wrapped in a smart pointer allocation.
(Allocating memory inside a function and then expecting the caller to free it again is a prime source for memory leaks.)
Don't write errors to cout, throw an exception.
Use nullptr, not NULL.

In short, write C++, not C with classes -- that was thirty years ago.

@AbdulMateen: Adding to the gratuitous generic advice: Don't change the question after posting it. Amended my answer to cover the added question. — DevSolar, Mar 01 '16 at 16:34
@AbdulMateen: One more -- [get a good book](http://stackoverflow.com/questions/388242/). No, really. ;-) — DevSolar, Mar 01 '16 at 16:35
Thanks you for your kind help and advice. I'll surely try to follow these. Sorry for my mistakes (I will try to improve) — Abdul Mateen, Mar 01 '16 at 16:38

NathanOliver · Answer 2 · 2016-03-01T16:34:56.597

1

It is possible to reach the end of your functions and you do not return anything at the end. If you have a function that is suposed to return something then every path out of the function needs to have a return. Basically

int * subArr(int pos, int siz)
{
    if (pos+siz < len)
    {
        subArr1 = new int[siz];
        for (int i=0; i<siz; i++)
        {
            subArr1[i] = arr[pos];
            pos++;
        }
        return subArr1;
    }
}

Becomes

int * subArr(int pos, int siz)
{

}

if if (pos+siz < len) is false. Since there is no return you get the error.

You second issue is you cannot overload a function based on its return type alone.

Array subArr(int pos, int siz)
int * subArr(int pos, int siz)

Ignoring the return type on the function we have

subArr(int pos, int siz)
subArr(int pos, int siz)

How is it supposed to tell them apart? It can't. That is why it is not allowed.

edited Mar 01 '16 at 16:34

answered Mar 01 '16 at 16:24

NathanOliver

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kindly check the question again Sir – Abdul Mateen Mar 01 '16 at 16:32
So it means that overloading depends upon the data type of arguments and not return type? – Abdul Mateen Mar 01 '16 at 16:36
@AbdulMateen Kindly ask the correct question. It is not polite to chamnge your question after you get an answer. I have updated. – NathanOliver Mar 01 '16 at 16:36

Writing c++ "array" class myself

2 Answers2