Question

i have to define a variable that i am using in a program but i want the variable to change from int to char.

example

#include <iostream>
#include <cstdlib>

int main()
{
    int l=rand();
    char x;
    std::cout<<"this program makes a random char or number enter n for a number or c if you want a letter\n";
    std::cin>>l>>"\n";
    if (x="c")
    {
        char l=rand();
        std::cout<<"here is your letter :"<<l<<"\n";
    }
    else if (x="n")
    { 
        int l=rand();
        std::cout<<"here is your number :"<<l<<"\n";
    }
}

so i already know there is other problems this was just an example i wrote real quick to show what i meant. i want to make it so that depending on what the user enters it will change l from a char to an int. I don’t know if there is something to do this but if there is i would like to know.

Thank You

Answer 1

The generic name for this kind of thing is a union.

You can use the union keyword to define a C-style union: a variable which may be treated as having one of several types. However, you need to do all the work to keep track of which type it is.

union {
    char c;
    int i;
} l;

bool l_is_int;

if ( x == 'c' ) {
    l_is_int = false;
    cin >> l.i;
} else if ( x == 'n' ) {
    l_is_int = true;
    cin >> l.c;
}

There is a class boost::variant which does this more elegantly and safely:

boost::variant< char, int > l;
if ( x == 'c' ) {
    char c;
    cin >> c;
    l = c; // set content type of l to char
    cout << "your char is " << boost::get< char >( l ) << endl;
} else {
    int i;
    cin >> i;
    l = i; // set content type of l to int
    cout << "your int is " << boost::get< int >( l ) << endl;
}
cout << "your int or char is " << l << endl;

---

For your immediate problem, using any kind of union is overkill, since an int can hold any char value:

int l;
bool l_is_int;

if ( x == 'c' ) {
    char c;
    cin >> c;
    l = c;
    l_is_int = false;
} else if ( x == 'n' ) {
    cin >> l;
    l_is_int = true;
}