重载多个运算符

Question

Concisely, my goal is to have foo[bar] return type1, and foo[bar]= return type2.

I am writing an object in C++, and it's coming along quite nicely, however there's just one tiny little thing that I wish to do, yet it seems impossible.

My object is a storage class, so I am using an array subscript to access values. I also need to assign, so I also overload the = operator. However, it is somewhat inconvenient because the values that my class holds are first class objects, so for my array subscript overload I can't return them as is. I have to return an intermediate class to handle the = operator, but I also want to retrieve the value without additional typing.

Is there any way to do this? Hackish ways are acceptable.

Edit: Here's an example of what it (should) do

#include<cstdio>
#include<cstdlib>

class foo{
    char* a[100];
    foo(){
        for( int i = 0; i < 100; i ++)
            a[i] = 0;
    }
    char* operator[] (int location){
        return a[location];
    }
    foo& operator[]= (int location, const char* value ){
        if( a[location] == 0 )
            a[location] = (char*) malloc( strlen( value ) + 1 );
        else
            a[location] = (char*) realloc( a[location], strlen( value ) + 1 );
        strcpy( a[location], value );
    }
};
int main(){
    foo bar;
    bar[20] = "Hello";
    printf( "bar[20] = %s\n", bar[20] );
    bar[20] = "Hello There";
    printf( "bar[20] = %s\n", bar[20] );
    printf( "bar[20][0] = %c\n", bar[20][0] );
}

Output:
bar[20] = Hello
bar[20] = Hello There
bar[20][0] = H

Edit again: I think I'll try phrasing this in a different, but workable way. Is there a way to overload the return type when a class is referenced? Such that if I have

class foo{
    bool a;
    bool operator /*referenced*/ (){
        return a
    }
    foo& operator=(bool b){
        a = b;
    }
};
int main(){
    foo a;
    a = b;
    if( a == b )
        printf("It Works!");
}

that would actually work?

Answer 1

There is no operator[]= , so the solution is to write some sort of wrapper class with two key features: an operator= that takes a value and sets it to the parent container, and an implicit conversion operator that takes the value from the parent container and returns it. Your operator[] will then return such wrapper.

class foo
{
    friend class wrapper;
public:
    class wrapper
    {
        friend class foo;
        foo & _parent;
        int _index;

        wrapper(foo & parent, int index) : _index(index), _parent(parent) {}
    public:  
        wrapper & operator=(const char * value)
        {
            if( _parent.a[_index] == 0 )
                _parent.a[_index] = (char*) malloc( strlen( value ) + 1 );
            else
                _parent.a[_index] = (char*) realloc( _parent.a[_index], strlen( value ) + 1 );
            strcpy( _parent.a[_index], value );

            return *this;
        }

        operator char *()
        {
            return _parent.a[_index];
        }
    };

    char* a[100];
    foo()
    {
        for( int i = 0; i < 100; i ++)
            a[i] = 0;
    }
    wrapper operator[] (int location){
        return wrapper(*this, location);
    }
};

For the second question, well, you could always overload operator== on foo . But maybe I misunderstood.

Answer 2

If you're willing to use C++ (as your tag suggests), then most of the work has been done for you:

class foo: public vector<string>
{
public:
  foo()
  {
    resize(100);
  }
};

int main()
{
   foo bar;

   bar[20] = "Hello";
   cout << "bar[20] = " << bar[20] << endl;
   bar[20] = "Hello There";
   cout << "bar[20] = " << bar[20] << endl;
   cout << "bar[20][0] = " << bar[20][0] << endl;
}