Problem with const array changing c++

Question

I have this code that tries to protect the user from array boundary errors.

I don't get why this will compile, tho i've declared the array to be const, therefore, i'm suppose to get a compilation error!

thanks a lot.

/************ file: SafeAccessArray.h ********************/
template<typename T>
class SafeAccessArray
{
private:
int _len;
T * _arr;
public:
SafeAccessArray (int len=2) : _len (len), _arr (new T [len]) {}
~SafeAccessArray () { delete _arr; }
T& operator [](int i) const
{if (i < 0 || i >= _len) throw (-1);
else return _arr[i]; }
};
/************ end of file: SafeAccessArray.h *************/

/************ file: SafeAccessArray1.cpp *************/
#include "SafeAccessArray.h"
int main()`enter code here`
{
SafeAccessArray<int> intArr (2);
intArr[0] = 0;
intArr[1] = 1;
const SafeAccessArray<int> intArrConst (2); // THIS IS THE "PROBLEMATIC" LINE
intArrConst [0] = 0;
intArrConst [1] = 1;
return 0;
}
/************ end of file: SafeAccessArray1.cpp ******/

SafeAccessArray needs a copy constructor and assignment operator. Look up the Rule of Three, and the copy-and-swap idiom. Even if this were a good idea, just use std::vector privately and be done with it. Also, don't throw integers, throw things deriving from std::exception; std::out_of_range awaits. — GManNickG
– GManNickG, Commented Oct 2, 2010 at 19:14
Please don't throw -1, not even in test code. Throw ::std::out_of_range or something, anything but a bare base type like int or char *, and preferably something derived from ::std::exception. — Omnifarious
– Omnifarious, Commented Oct 2, 2010 at 19:21
In case you're using G++/libstdc++, just use std::vector, compile with -D_GLIBCXX_DEBUG and enjoy libstdc++'s debug mode. Other compiler vendors probably offer a similar feature. — sellibitze
– sellibitze, Commented Oct 2, 2010 at 19:28

GManNickG · Accepted Answer · 2010-10-02 19:14:11Z

4

Yea it's const, but you did T& operator [](int i) const anyway. You're returning a reference, and this function can be called on a const object.

Make it return const T&. Better yet, stop. Just use std::vector and the at() function.

answered Oct 2, 2010 at 19:14

GManNickG

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4 Comments

limlim Over a year ago

so, just to make it clear, when i call "intArrConst [0] = 0" - operator[] will cause a cast to a non-const object, or the assignment won't really happen on my const object?

GManNickG Over a year ago

@limlim: What? When you say intArrConst[0], T& operator [](int i) const is called. That results in a T&, and then you assign 0 to it, changing the reference (which, being an alias, just changes the value it's aliasing.)

Potatoswatter Over a year ago

@limlim: The class contains T * _arr; a const instance of the class contains a T *const _arr whereas you want it to have a T const *const _arr. C++ did not add the extra qualification you wanted, so the guarantee is your responsibility.

GManNickG Over a year ago

@jenny: If an answer solved your problem, please check the checkmark next to it.

Arun · Accepted Answer · 2010-10-02 20:26:57Z

2

I think that operator[] member function desires the following two overloaded variants:

T& operator [](int i);
const T& operator [](int i) const;

The one provided

T& operator [](int i) const;

does not match any of the above, and hence the problem.

answered Oct 2, 2010 at 20:26

Arun

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