Variable name with using pointer

Question

I want to pick from keyboard which array i want to display (i know my code doesn't work i just want to show my problem)

 int main(){
            char *a = new char[5];
            char *b = new char[5];
            char name;

            cin >> a;
            cin >> b;
            cin >> name;

            cout << name; 
}

i will have char arrays from a-z is there other way to do this with using pointers or something :/ — Rooster, Apr 03 '19 at 18:32
In the future you may receive better feedback by telling us a bit more of what you have tried and how it failed specifically :) — Pablo, Apr 03 '19 at 18:33
@Rooster a better way to say *i can't use stl* is *I can't use library containers*. Otherwise you get response like above, or *No one said anything about STL. We're talking the C++ Standard Library here.* or *Like... Omigawd! STL is, like, soooooo 1990s* — user4581301, Apr 03 '19 at 19:03

463035818_is_not_an_ai · Answer 1 · 2019-04-03T18:43:51.753

The variable names you use in your code have no meaning once your program is running. You cannot dynamically replace name with either a or b, if thats what you wanted to do.

Making the example simpler (c-style arrays are definitely not for beginners, look at std::vector instead), you can do this:

#include <iostream>
int main() {
    int a = 42;
    int b = 102;
    std::cout << "select a or b: ";
    char select;
    std::cin >> select;
    if (select == 'a') std::cout << a;
    else if (select == 'b') std::cout << b;
    else std::cout << "wrong input \n";
}

score 0 · Answer 2 · answered Apr 03 '19 at 18:50

Most of the time, if you want to have a number of objects (say, more than 2 or 3) of the same type, you should consider using an array or vector.

So instead of

char *a = ...;
char *b = ...;
char *c = ...;

Use

char *values[] = { ... };

Now you can access a value by using an index, i.e. an integer value. For a character a-z, this can easily be done by subtracting the value 'a' ('a'-'a' is 0, 'b'-'a' is 1, etc).

const char *values[] = { "this is a", "this is b" };

char name;
std::cin >> name;
int index = name - 'a';

// TODO: Make sure the index is in range!
const char *value = values[index];
std::cout << value;

Variable name with using pointer

2 Answers2