Using Array as parameter in function in C++ - c++

Question

This program gets two hexadecimal numbers and converts them to decimal numbers. And it finally returns sum of two number in decimal form.
Before I enter "num2", "n1" gets right value.
But "n1" becomes 0 after I get "num2".
I don't know why this happens.
Please tell me the reason in hurry...
#include <iostream>
using namespace std;
class HextoDec {
public:
void getNum();
int add();
private:
char num1[2], num2[2];
int convert(char num[]);
int n1, n2;
};
int main()
{
HextoDec a;
a.getNum();
cout << "Sum of two number is " << a.add() << endl;
}
void HextoDec::getNum() {
cout << "Enter first number : ";
cin >> num1;
n1 = convert(num1);
cout << "Enter second number : ";
cout << endl << n1 << endl; // Value of n1 is correct
cin >> num2;
cout << n1 << endl; // Problem occurs. Value of n1 becomes 0
n2 = convert(num2);
}
int HextoDec::convert(char num[]) {
int j = 16, n = 0;
for (int i = 0 ; i < 2 ; i++) {
switch (num[i]) {
case '0':
n = n + j * 0; break;
case '1':
n = n + j * 1; break;
case '2':
n = n + j * 2; break;
case '3':
n = n + j * 3; break;
case '4':
n = n + j * 4; break;
case '5':
n = n + j * 5; break;
case '6':
n = n + j * 6; break;
case '7':
n = n + j * 7; break;
case '8':
n = n + j * 8; break;
case '9':
n = n + j * 9; break;
case 'A':
n = n + j * 10; break;
case 'B':
n = n + j * 11; break;
case 'C':
n = n + j * 12; break;
case 'D':
n = n + j * 13; break;
case 'E':
n = n + j * 14; break;
case 'F':
n = n + j * 15; break;
}
j = 1;
}
return n;
}
int HextoDec::add() {
cout << "*****" << endl;
cout << n1 << endl;
cout << n2 << endl;
return n1 + n2;
}
What's the reason?
What makes this happens?
What can I do or should I do to solve this problem?

Answer 1

As other people have mentioned, your char arrays only contain one element, plus the '\0' character. If you try to read with cin a two character array (i.e. 1a) you will have a undefined behaviour. This example might give you some hints about '\0'
const char *example1 = "hi";
strlen(example1); // This is 2
const char *example1 = "hi\0";
strlen(example1); // This is also 2
But answering to your question, if you only want to read 2-digit hexadecimal values, char num1[3] should fix the issue.
However, I think you might have another issue. Just copied/pasted your code and if I enter just 1 the result that I get is 16, but it should be 1. Maybe you expect the user to input 01 instead.
I guess this is a didactic example, so you might want to play a bit with the code, and for example be able to convert any number to decimal. Maybe you can modify your function convert with something like:
int HextoDec::convert(char num[]) {
int n = 0;
int sizeNumber = strlen(num);
for (int i = sizeNumber - 1; i >= 0; i--)
{
int j = pow(16, sizeNumber - i - 1);
...
}