Accessing Member Variables of Derived Structs Without Casting - c++

Question

I have a bunch of derived structs. Is there any way for me to access the derived width without casting?
If I do this, I get 1s and not 42 and 13:
#include <iostream>
#include <deque>
using namespace std;
struct BaseParams {
virtual ~BaseParams() {}
int width = 1;
};
struct VectorParams : BaseParams {
int width = 42;
};
struct MatrixParams : BaseParams {
int width = 13;
};
int main()
{
std::deque<BaseParams *> params;
VectorParams *vectorParams = new VectorParams;
MatrixParams *matrixParams = new MatrixParams;
params.push_back(vectorParams);
params.push_back(matrixParams);
for (auto &param : params) {
std::cout << param->width << std::endl;
}
// Don't want to do this
// for (auto &param : params) {
// MatrixParams *matrixParams = dynamic_cast<MatrixParams *>(param);
// if (matrixParams)
// std::cout << matrixParams->width << std::endl;
// VectorParams *vectorParams = dynamic_cast<VectorParams *>(param);
// if (vectorParams)
// std::cout << vectorParams->width << std::endl;
// }
return 0;
}
If possible, I want to avoid casting because I have so many derived structs.

Answer 1

To overwrite the width in BaseParams, instead of making a new variable, you can give VectorParams a constructor like this:
struct VectorParams : BaseParams {
VectorParams() {
width = 42;
}
};
In fact constructors are always how variables are set in new objects. When you write
struct BaseParams {
virtual ~BaseParams() {}
int width = 1;
};
the compiler actually turns it into a constructor, similar to
struct BaseParams {
virtual ~BaseParams() {}
int width;
BaseParams() {
width = 1;
}
};