operator[] ? lvalues!?

[earlz]

How can I make an object so that I can go MyObject[5]=93;

it says that operator[] takes exactly 1 argument, so how the crap is it possible?

[Colonel Kernel]

How can I make an object so that I can go MyObject[5]=93;

it says that operator[] takes exactly 1 argument, so how the crap is it possible?

Return a non-const reference to the item that you want to receive the value 93. The 1 argument is the index -- 5 in your example. The assignment happens after operator[] returns.

[Candy]

How can I make an object so that I can go MyObject[5]=93;

it says that operator[] takes exactly 1 argument, so how the crap is it possible?

int &operator[](int index);

[earlz]

It doesn't really make sense to me, but it works...so...

[os64dev]

How can I make an object so that I can go MyObject[5]=93;

it says that operator[] takes exactly 1 argument, so how the crap is it possible?

int &operator[](int index);

ahum.. because none of the members change it should be:

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int &operator[](int index) const;

[Colonel Kernel]

How can I make an object so that I can go MyObject[5]=93;

it says that operator[] takes exactly 1 argument, so how the crap is it possible?

int &operator[](int index);

ahum.. because none of the members change it should be:

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int &operator[](int index) const;

If you take "const" literally, yes, but it is not always a good idea to do so. Think about the bigger picture. Do you really want to allow code like this?

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const MyClass MyObject; // Should NEVER change!
MyObject[5] = 93; // Whoops, we made operator[] const and let something change...

[os64dev]

ok but f you take "const" literally, yes, but it is not always a good idea to do so. Think about the bigger picture. Do you really want to allow code like this?

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const MyClass MyObject; // Should NEVER change!
MyObject[5] = 93; // Whoops, we made operator[] and let something change...

maybe i misunderstood but:

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int &operator[](int index) const;

just tells the compiler that this member function does not change the content of the current object and can generate more efficient code. The const here does not specifie a constant return value or a constant object. The code example you gave will not compile.

[Solar]

maybe i misunderstood but:

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int &operator[](int index) const;

just tells the compiler that this member function does not change the content of the current object and can generate more efficient code.

Actually it tells the compiler that this function is valid for const objects. Without the trailing const in the function declaration, a function can only be called for non-const objects.

Do you really want to allow code like this?

Do you really want to disallow code like this (which you would if you did not declare the operator const)?

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const MyClass myObject( ... );
cout << myObject[5];

Depending on the actual semantics of MyClass, you could write something like this to disallow the code you objected to:

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class MyClass {
    public:
        MyClass const & operator[]( int index ) const;
        MyClass & operator[]( int index );
};

That way, the reference returned by operator[]() is const if the MyClass object is const, and non-const otherwise.

[os64dev]

hmm, it seems i have to re-read the c++ book again.

EDIT: me confused.

constant member function

[Colonel Kernel]

Do you really want to disallow code like this (which you would if you did not declare the operator const)?

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const MyClass myObject( ... );
cout << myObject[5];

Of course not.

Depending on the actual semantics of MyClass, you could write something like this to disallow the code you objected to:

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class MyClass {
    public:
        MyClass const & operator[]( int index ) const;
        MyClass & operator[]( int index );
};

That way, the reference returned by operator[]() is const if the MyClass object is const, and non-const otherwise.

Exactly.

Next time I will endeavour to tell more than just half the story.

[Candy]

ahum.. because none of the members change it should be:

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int &operator[](int index) const;

Actually, if you return a reference to a pointer-contained object in this object, this would compile for const objects and be wrong - since you do change the state (or by const-capable actions you perform allow const object visible behaviour to change). Use the method Solar suggested - make two, one for const and one for non-const. Oh, and if you still use MSVC, kick it. VS2005 can't distinguish between const and non-const in at least the copy constructor, so I'm somewhat doubting it can do the other cases.

constant member function

I refute your attempt to pass MSDN as C++ documentation. Not to mention the non-HTML compatible link.

[os64dev]

Oh, and if you still use MSVC, kick it. VS2005 can't distinguish between const and non-const in at least the copy constructor, so I'm somewhat doubting it can do the other cases.

i use a gcc/g++ cross compiler so no issues here

I refute your attempt to pass MSDN as C++ documentation. Not to mention the non-HTML compatible link.

lol

[AndrewAPrice]

You "could" do it the manual way.

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myObject *myArray;

int NumberOfItems = 5;

myArray[] = new char[NumberOfItems * sizeof(myObject)];
for(int i = 0; i < NumberOfItems; i++)
    myArray[i].myArray(); // call the constructor

then to clean up:

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for(int i = 0; i < NumberOfItems; i++)
    myArray[i].~myArray(); // call the deconstructor
delete myArray;

[Candy]

You "could" do it the manual way.

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myObject *myArray;

int NumberOfItems = 5;

myArray[] = new char[NumberOfItems * sizeof(myObject)];
for(int i = 0; i < NumberOfItems; i++)
    myArray[i].myArray(); // call the constructor

then to clean up:

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for(int i = 0; i < NumberOfItems; i++)
    myArray[i].~myArray(); // call the deconstructor
delete myArray;

And what exactly, aside from a load of code and a high complexity, does this add over

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myObject myArray[5];

?

[AndrewAPrice]

In my way you could unwrap the loop that calls the constructors so that each array element can be passed individual constructor parameters.