[Solved] Why doesn’t the char convert to an int?

Question

There are three things confusing you here.

First the expressions:

i1 = data[0];
i2 = data[p1];

Assign the character values of data[0] and data[p1]
If the input is 2+4 you’re actually assigning the character values of ‘2’ and ‘+’. In ASCII “2” is represented by code-point 50 and “+” by 43.

So then the expression i1+i2 actually equals 93.

What is also confusing you is that s=i1+i2 actually binds to the overloading basic_string& operator=( CharT ch ); and so converts 93 back to a character and assigns a singleton string to s which happens to be ] as you see!

This is your code with those three faults fixed:

#include <iostream>

using namespace std;

int main()
{
    string input;
    int length;
    int p1;
    int i1;
    int i2;
    string result;

    cout << "Enter your expression" << endl;
    cin >> input;
    length = input.size();

    char data[length];
    input.copy( data, length );

    switch(data[1]){
    case '+' : p1 = length - 1; //Not -2 as was.
    i1 = data[0]-'0';//Works for ASCII compatible and so all modern systems...
    i2 = data[p1]-'0';
    result = std::to_string(i1 + i2);//Force the creation of a decimal string of the result.
    cout << result << endl;
    break;
    }
    return 0;
}

Accepted Answer