declaring a variable with a specified type in c++

Initially, this seems like very trivial problem.

I have input: string a, string b.
b only takes two values, "string" and "int".

I want to declare a new variable var0 with a type that corresponds to b.

How to implement the logic?

I thought this function would work, but it won't.

auto varx(std::string a, std::string b){  
    if(b == "int") return std::stoi(a);  
    return a;  
}

e.g. auto var0 = varx("43", "int"); should set var0 as int var0 = 43;

I have seen union or variant in discussions, but I'm not sure how to implement it.

You can do that using the desired functionality in C++, you can use std::variant to handle the different types dynamically. Here is a snippet:

#include <iostream>
#include <variant>
#include <string>

// Define a variant that can hold either an int or a string
using VariantType = std::variant<int, std::string>;

VariantType varx(const std::string& a, const std::string& b) {
    if (b == "int") {
        return std::stoi(a);  // Convert the string to an int
    }
    return a;  // Return the string as is
}

int main() {
    VariantType var0 = varx("43", "int");
    VariantType var1 = varx("Hello", "string");

    // To get the value out of the variant, you need to use std::get
    if (std::holds_alternative<int>(var0)) {
        std::cout << "var0 is an int: " << std::get<int>(var0) << std::endl;
    } else {
        std::cout << "var0 is a string: " << std::get<std::string>(var0) << std::endl;
    }

    if (std::holds_alternative<int>(var1)) {
        std::cout << "var1 is an int: " << std::get<int>(var1) << std::endl;
    } else {
        std::cout << "var1 is a string: " << std::get<std::string>(var1) << std::endl;
    }

    return 0;
}