2

Why the following code is giving me an error

Test.cpp:23:10: error: invalid conversion from ‘const int*’ to ‘int*’ [-fpermissive] return array;

#include <iostream>
#include <stdio.h>

#define MAX_ELEMENTS 5
class CBase
{

public:
    CBase()
    {
        for(int i = 0; i < MAX_ELEMENTS; i++)
        {
            array[i] = 0;
        }
    }
    ~CBase()
    {
        // Nothing
    }
    int * GetArray() const
    {
        return array;
    }

private:
    int array[MAX_ELEMENTS];

};

int main ()
{
    CBase b;
    return 1;
}

EDIT: I understand that I should return a const int * but then I tried something below which works fine, request to explain the reason for allowing this and not allowing the above.

#include <iostream>
#include <stdio.h>

class CBase
{
public:
    CBase():ptr(NULL)
    {
    }
    ~CBase()
    {
        delete ptr;
    }
    int * ptr;
public:
    int * GetPtr() const
    {
        return ptr;
    }
};

int main ()
{
    CBase b;

    return 1;
}
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8
  • It really does break const-correctness when you let the caller modify a const object's data. Commented Feb 4, 2014 at 17:52
  • Try const int * GetArray() const {...} Commented Feb 4, 2014 at 17:52
  • because your const method should return int const* and not int* Commented Feb 4, 2014 at 17:52
  • but then how const member function returning plain variable as it is (i.e. non array) without any error? Commented Feb 4, 2014 at 17:55
  • @user3271978, It works exactly the same way. You don't have to return a const int * const, just like you don't have to return a const double or something. The return value is copied out. Commented Feb 4, 2014 at 17:56

3 Answers 3

Reset to default
5

Imagine code like this:

const CBase b;
int *array = b.GetArray();
array[0] = 5; // ooops! b changed but we declared it const!?

So as already mentioned in the comments, it does break const-correctness of your code. What you need to do is either declare GetArray() non-const, or make it return a pointer to a const int.

const int * GetArray() const 
{
    return array;
}

Now, code like this would not compile:

const CBase b;
const int *array = b.GetArray();
array[0] = 5;

EDIT To answer your other question from the comments above: When you call a method that returns a value and you assign this return value to some variable, than the return value is copied to your variable and afterwards discarded. So when your calling code changes the value of this variable this has no effect on the value or variable that was initially returned. That is why a const member function can return some of the class' data members. However, when you return a pointer to a data member, than the calling code can manipulate the value of this member. Still, the pointer is copied, but even the copy points to the memory location where the class member is stored and thus you could manipulate its value.

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1 Comment

I think your example would be better if b was const.
2

Your method should return a const int* instead.

const int * GetArray() const
{
    return array;
}
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Comments

0

Its simple you need to either declare GetArray() non-const, or make it return a pointer to a const int.

const int * GetArray() const
{
    return array;
}

The reason is that if you return non constant array then as array is returned as pointer so its value can be changed by function which gets its value so constant function indirectly results in changing the value so you need to return constant array.

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