Re: Returning an unknown number of types/values
From: Jumbo (nospam)
Date: 01/16/04
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Date: Fri, 16 Jan 2004 22:30:20 -0000
"Tony Burrows" <tony@tonyburrows.com> wrote in message
news:pan.2004.01.15.09.51.54.364880@tonyburrows.com...
> Thanks for all the help! I'll try some of these.
>
> I had actually got to the idea of vector< vector< tObject *> > where my
> tObject is your Base, but I'd subclassed to all the possibilities and then
> had to use a dynamic cast when getting the value back. Overloading get is
> so much better.
>
> I'd never heard of boost, now to investigate that.
>
Perhaps my code will make it clearer what I am trying to explain:
#include <iostream>
#include <string>
#include <vector>
using std::cout;
class Base{
public:
virtual ~Base()=0{}
virtual void setData(int){}
virtual void setData(double){}
virtual void setData(std::string){}
//virtual void setData(){}
};
class Variant{
public:
Variant():itsArray(0),itsLen(0){}
Variant(const std::vector<std::string> &v1)
:itsArray(new Base*[v1.size()]),
itsLen(v1.size()),
itsTypes(v1){
for(size_t i=0;i<v1.size();i++){
if(v1[i]=="int")itsArray[i] = new DataType<int>;
else if(v1[i]=="double")itsArray[i] = new DataType<double>;
else if(v1[i]=="std::string")itsArray[i] = new DataType<std::string>;
else itsArray[i] = 0;
}
//itsArray[3]->setData();??
}
~Variant(){
for(size_t i=0;i<itsLen;i++){delete itsArray[i];}
delete itsArray;
}
size_t getLen()const{return itsLen;}
Variant(const Variant& rhs):itsLen(rhs.getLen()),itsArray(new
Base*[itsLen]){
for(size_t i=0;i<itsLen;i++){
if( rhs.itsTypes[i]=="int")itsArray[i] = new DataType<int>;
else if(rhs.itsTypes[i]=="double")itsArray[i] = new DataType<double>;
else if(rhs.itsTypes[i]=="std::string")itsArray[i] = new
DataType<std::string>;
else itsArray[i] = 0;
}
}
Variant& operator=(const Variant& rhs){
itsLen = rhs.getLen();
itsArray = new Base*[itsLen];
for(size_t i=0;i<itsLen;i++){
if( rhs.itsTypes[i]=="int")itsArray[i] = new DataType<int>;
else if(rhs.itsTypes[i]=="double")itsArray[i] = new DataType<double>;
else if(rhs.itsTypes[i]=="std::string")itsArray[i] = new
DataType<std::string>;
else itsArray[i] = 0;
}
return *this;
}
private:
std::vector<std::string> itsTypes;
size_t itsLen;
Base** itsArray;
template<typename T>
class DataType: public Base{
public:
DataType():data(0){cout<<"Default DataType constructor"<<'\n';}
~DataType(){cout<<"DataType dest..."<<'\n';}
void setData(){cout<<"setting data for type: "<<typeid(T).name()<<'\n';}
private:
T data;
};
};
/* Specialize any types you don't want initialised to 0 */
/* Specialized std::string constructor */
template<>
Variant::DataType<std::string>::DataType():data("0"){
cout<<"DataType const with std::string"<<'\n';
}
int main(){
std::vector<std::string> strArr1(4);
strArr1[0] = "int";
strArr1[1] = "std::string";
strArr1[2] = "double";
strArr1[3] = "jtd::jing"; /*deliberate error*/
std::vector<std::string> strArr2(3);
strArr2[0] = "std::string";
strArr2[1] = "std::string";
strArr2[2] = "double";
Variant v1(strArr1); /*Will have 4 Datatypes, one will be null.*/
Variant v2(strArr2); /*Will have 3 Datatypes.*/
std::vector<Variant> vv(2);
vv[0] = v1;
vv[1] = v2;
// etc etc....
return 0;
}
This class works with vectors and I am currently looking into how it works
with iterators, algorithms and maps and I think I can make it fully
compatable with the complete standard library.
I haven't posted it with accessor functions etc as I don't know what you
need. Do you want to initialise the Variant DataTypes with a array of
strings or do you want to initialise each piece of data individually and how
do you want to access the data etc. I have included a basic foundation for
setter functions and from this it should be easy to understand how to
implement any type of accessor methods.
HTH.
Note : You may encounter problems if you think you can overload the
subscript( [] )operator to return a particular index because this would mean
the function only differs return type, which I don't think is allowed in
function overloading(please correct me if I am wrong). However think I may
have worked out a way in which this can be done but I have yet to test it.
- Next message: Richard Heathfield: "Re: working with addresses"
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