table = new myObject*[TABLE_SIZE];
I know that * are for declaring pointers variable and to obtain the value of the variable from a pointer, but what does this mean?
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Note that this is typically bad C++ style. A better solution would be `std::vector
table(TABLE_SIZE, nullptr)`. But I suspect there are more problems, such as the use of non-smart pointers and possible a ` #define TABLE_SIZE`. – MSalters Nov 08 '16 at 11:11
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It means exactly that.
table is a pointer to an array (with size TABLE_SIZE) of pointers to objects of type myObject. Note that you haven't allocated any memory for those pointers at this point.
Don't forget to call delete[] table; once you're done with it.
Bathsheba
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... or better, just replace with `std::vector
table(TABLE_SIZE);` and stop having to worry about calling `delete []`. – Martin Bonner supports Monica Nov 08 '16 at 10:27
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This means that you allocate an array of type myObject* of size TABLE_SIZE.
So in this case we allocate space for the pointers but not the objects.
Your variable table will then be a pointer to an array of pointers, or a myObject**.
So table will point to the first element in this list of pointers.
Tommy Andersen
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The asterisk means pointer here too. I am assuming that table is declared like
myObject** table;
What you are doing is allocating an "array" of TABLE_SIZE pointers to myObject.
Some programmer dude
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It means exactly what you already know it means :-)
The type myObject * is a pointer to a myObject object so what you are declaring is an array of said pointers, TABLE_SIZE of them to be exact.
paxdiablo
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