How to output a variable stored in another variable - c++

Question

So here's my code at the moment:
#include <iostream>
using namespace std;
int main() {
string x = "_one";
int sound_one = 7;
int sound_two = 8;
cout << ("sound") + x;
}
However when I run the code, it outputs 'sound_one' instead of '7'. How do I get it to output the variable sound_one instead of just 'sound_one'? Also, I need it so I can change x to different things (eg '_two') so it will then output sound_two instead. Any help would be greatly appreciated.

Answer 1

C++ is not a reflective language in the sense that you can acquire a variable name at runtime (variable names are normally compiled out of the program). You can use std::map though to achieve your immediate aim:
#include <iostream>
#include <string>
#include <map>
int main() {
using std::literals::string_literals::operator""s;
std::string x = "_one";
std::map<std::string, int> data;
data["sound_one"] = 7;
data["sound_two"] = 8;
std::cout << data["sound"s + x];
}
Note the notation "sound"s: the suffixed s denotes a std::string user defined literal.

Answer 2

You can't a variable from a string in this way. A work around is to use if/switch statements until the variable name is matched and then print it:
if(x == "_one") {
cout << sound_one;
}
else if(x == "_two") {
cout << sound_two;
}
else {
cout << "no match";
}

Answer 3

You can't do that in C++. You can use a map as shown in another answer, but I feel like what you really need is an array:
#include <format>
#include <iostream>
int main()
{
int data[] { 7, 8 };
std::cout << std::format("data[{}] = {}\n", 0, data[0]);
}
Arrays are usually better for such simple ordered sequences.