Reputation: 31
I need to build a vector of a class that can have multiple type like this:
#include <variant>
#include <vector>
#include "Field.h"
using namespace std;
int main()
{
variant <int, float> v;
vector <variant<Field<int>, Field<string>, Field<float>>> fdList;
fdList[0].getName();
}
And this is header file Field.h
:
#pragma once
#include <string>
#include <vector>
using namespace std;
template<class T>
class Field
{
public:
Field();
Field(string);
void setName(string);
string getName();
bool isPrime();
void toPrime();
void toForeign(Field);
~Field();
private:
string FD_Name;
vector <T> records;
bool isPrimeK = false;
string message;
};
template<class T>
string Field<T>::getName()
{
return FD_Name;
}
When I try to access getName()
function, Visual Studio keeps giving me the following message error:
E0135 class "std::variant<Field, Fieldstd::string, Field>" has no member "getName"
C2039 'getName': is not a member of 'std::variant<Field,Fieldstd::string,Field>'
But it works just fine, if I define my vector like this:
vector <Field<int>> fdList;
fdList[0].getName();
How can I fix this?
Upvotes: 2
Views: 4962
Reputation: 7904
Calling a method on variant
doesn't automatically call the method of the active alternative of variant
. You'll have to visit
the active alternative and invoke the corresponding handler. In your case, since you want to handle all of the potential active alternatives the same way, you can do:
std::visit([](const auto& field) {
field.getName();
// ...
}, fdList[0]);
Alternatively, you can wrap the variant
in something like:
struct AnyField {
string getName() const {
return std::visit([](const auto& field) { return field.getName(); }, v);
}
std::variant<Field<int>, Field<string>, Field<float>> v;
};
then use them like you wanted to:
vector<AnyField> fdList;
fdList[0].getName();
Upvotes: 1
Reputation: 586
For any issue about standard library, I recommend you to check document first.
You can see here about how to use std::variant
.
In short, you cannot access the content in your std::variant
like that because its type is std::variant
but not the types you store in it. For your case, I think you may want to check what's inside first by calling std::variant::index()
method, then get the value by calling std::get
.
Upvotes: 1