Printing out chars as letters without turning them into numerical values?

Question

My problem is simple but I have no idea on solving it.

I have an integer variable which gets random numbers. But on special occasions this integer represents chars. For instance, if rand() function gives 100, the variable becomes A. I used switch-case to manage these exceptions. Then I executed the program and rand() gave 100 for a variable. However, instead of printing out A char, program gave its ASCII value, 65. I thought about opening up another print function and implementing type casting for these exceptions. But there's a lot of them, and also, I'm getting a lot of random numbers during the program, so it's almost impossible to make this happen.

// Program gets random values for a lot of variables (`rand_val1`, `rand_val2`, `rand_val3`...)

// For some integers, it converts them to pre-defined chars

char A = 'A', char B = 'B' ...

...

switch (rand_val1)
   case 100:
   rand_val1 = A;
   break;

   case 200:
   rand_val2 = B;
   break;

   ...

switch (rand_val2)

   ...

// It prints out each one of them.

cout << rand_val1 << " " << rand_val2 << ... << endl;

/* As output, it doesn't give chars and instead it gives their ASCII values

>>> 65 66 300 400 500 70 71 ...

What can I do in this case? Any ideas?

Answer 1

Try This one. This is a much simple If Else condition but based on your requirements, it does not need to be dynamic

#include<iostream>
using namespace std;

char Convert(int input)
{ char ToBeReturn = ' ';

if (input == 100)
{
    ToBeReturn = 'A';
}
if (input == 200)
{
    ToBeReturn = 'B';
}
if (input == 300)
{
    ToBeReturn = 'C';
}
if (input == 400)
{
    ToBeReturn = 'D';
}

return ToBeReturn;
}


void main()



{ int rand_val1, rand_val2, rand_val3;

cout << "Enter Value 1: ";
cin >> rand_val1;
cout << "Enter Value 2: ";
cin >> rand_val2;
cout << "Enter Value 3: ";
cin >> rand_val3;

cout << Convert(rand_val1) << " " << Convert(rand_val2) << " " << Convert(rand_val3) << endl; }

Answer 2

Because you told there are a lot of such conversion, so I dont think any function / std::cout type casting suit you. You can use a wrapper class here. Just declare all the rand_valXXX as Integer class.

If every rand_val have different conversion logic, then I think you have no way, but cast it every time when you print.

#include <iostream>

class Integer
{
public:
    Integer( int a ) : _a(a){}

    operator int() const { return _a; }

private:
    int _a;
};

std::ostream& operator<< ( std::ostream& out, const Integer& val )
{
    switch( val )
    {
    case 100:
        out << 'A';
        break;
    case 200:
        out << 'B';
        break;
    default:
        out << (int)val;
    }
    return out;
}

int main()
{
    Integer a = 100;
    std::cout << a;

    a = 200;
    std::cout << a;

    a = a + 100;
    std::cout << a;

    a = a / 200;
    std::cout << a;
}

Answer 3

How about defining an "exception list" ? Here's an example

#include <bits/stdc++.h>
using namespace std;

int main ()
{
    const unordered_map<int, char> exceptionList = {
        { 100, 'A'},
        { 200, 'B'},
        { 300, 'C'},
        { 400, 'D'},
        { 500, 'E'}
    };

    int rand_val1 = 100;

    if (exceptionList.find(rand_val1) == exceptionList.end())
    {
        cout << rand_val1 << endl;
    }
    else
    {
        cout << exceptionList.at(rand_val1) << endl;
    }

    return 0;
}

Or with a lambda:

#include <bits/stdc++.h>
using namespace std;

int main ()
{
    const std::unordered_map<int, char> exceptionList = {
        { 100, 'A'},
        { 200, 'B'},
        { 300, 'C'},
        { 400, 'D'},
        { 500, 'E'}
    };

    int rand_val1 = 100;
    int rand_val2 = 101;

    const auto charOrInt = [&exceptionList] (const int val) -> string {
        if (exceptionList.find(val) == exceptionList.end())
        {
            return to_string(val);
        }
        else
        {
            return string{exceptionList.at(val)};
        }
    };

    cout << charOrInt(rand_val1) << " " << charOrInt(rand_val2) << endl;

    return 0;
}