Question

Im having a little problem with classes.

In code:

class A
{
public:
     int num;
     sayNum() { cout<<num;}
};

class B : public A
{
public:
     sayNum() { cout<<"my num is: "<<num;}
};

/*somewhere else...*/
A foo[10];

B something;
something.num=10;

foo[0]=something;

foo[0].sayNum();
/*...*/

When I call foo[0].sayNum(); it prints “10”, and I would like it to print “my num is: 10”. I can’t change the type of the array. Note that it is just an example code, if I paste mine here it would be hard to understand :D)

Answer 1

This line:

foo[0]=something;

copies something into the array. Because the type of something (i.e. class B) is not the same as the type of foo[0] (i.e. class A), there is a conversion. In this case, something will be sliced, and only the A part will survive.

The net result is that foo[0] still contains an object with type class A, and thus the “my num is” will never be output.

Changing sayNum() to be virtual, which is good advice to solve other problems your code will show, won’t help in this case.

How to solve it, depends on what you want to achieve. Possible solutions are:

• Change the array to contain pointers: A* foo[10]; Note: this means you’ll have to use ‘new’ and ‘delete’.
• Use a std::vector instead of an array: std::vector<A*> foo; Additional benefit: a vector can grow and shrink on demand.
• Use smart pointers: std::vector<std::shared_ptr<A> > foo; Additional benefit: no worries about delete anymore.

But without knowing what you want to achieve, it is impossible to suggest the correct approach.