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It's possible to overload operator[] to take one argument, in the brackets. However how can I overload this operator to allow the user to type object[a] = b?

user3150201
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  • Of course, you decide whatever is returned, and whatever is returned by the operator can be assigned to – Gunther Van Butsele Oct 11 '14 at 18:00
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    Technically the way of speaking about it is that your operator needs to return an "[LValue](http://stackoverflow.com/questions/3601602/what-are-rvalues-lvalues-xvalues-glvalues-and-prvalues)"...which means it can appear on the left-hand side of an assignment. So basically return a non-const reference that can be a target of the kind of assignment you want to make, and is a meaningful place to be targeting said assignment. – HostileFork says dont trust SE Oct 11 '14 at 18:03

3 Answers3

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You return a reference to your item, something like:

A &operator[](int pos) { return list[pos]; }

So when you say object[a]=b; it's actually translated to:

A &tmp=object[a];
tmp=b;

Or in more clear terms:

A *tmp=object[a];          // more or less a pointer
*tmp=b;                    // you replace the value inside the pointer
Blindy
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1

The general declaration of the operator (it shall be a class member function) is the following

ObjectElementType & operator []( size_t );
Vlad from Moscow
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1

Subscript operators often comes in pair :-

T const& operator[] (size_t ) const;   
T&       operator[] (size_t );  // This enables object[a] = b on non-const object
P0W
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