Making a sum with char - c++

Question

#include <iostream>
#include <cstring>
using namespace std;
int test(char[], int);
void decal(char[], int n);
int main()
{
char a[10], b[10], c[10];
int valid;
do {
cout << "Insert first number (maximum 5 chars hexa):" << endl;
cin >> a;
valid = test(a, strlen(a));
if(!valid)
cout << "Error." << endl;
} while (!valid);
cout << "First number: " << a << endl;
decalez(a, strlen(a));
cout << "First number after insert: " << a << endl;
do {
cout << "Insert 2nd number (maximum 5 chars hexa):" << endl;
cin >> b;
valid = test(b, strlen(b));
if(!valid)
cout << "Error." << endl;
} while (!valid);
decalez(b, strlen(b));
cout << "2nd number after insert: " << b << endl;
add(a, b); // Calculating c
cout << "Result: " << c << endl;
return 0;
}
int test(char x[], int n)
{
if(n > 5)
return 0; // Too many numbers
for(int i = 0; i < n; i++)
{
if(x[i] <48 || (x[i] > 57 &&
x[i] < 65) || (x[i] > 70 && x[i] < 97) || x[i] >
102)
return 0;
}
return 1;
}
void decal(char x[], int n)
{
int i, nz;
x[5] = '\0';
nz = 5 - strlen(x);
if(nz > 0) {
for(i = 0; i < n; i++)
x[5 - i- 1] = x[n-i-1];
}
for(i = 0; i < nz; i++)
x[i] = '0';
}
I was given this school project to make a hexadecimal calculator. The teacher made us the following code. Mandatory
My problem is the void add part. How is it possible to add char?
I know there are easier ways to make a hexadecimal calculator, but we have to use that code.
So how can I write a sum something like 1cec+bec=28d8 in hexa?

Answer 1

Assuming the existence of these functions
// Convert hexadecimal string 'number' into an integer.
int fromHex(const char* number);
// Convert 'number' to hexadecimal in 'result', assuming that the result will fit.
void toHex(int number, char* result);
you can write add like this:
void add(const char* hex_a, const char* hex_b, char* hex_c)
{
int a = fromHex(hex_a);
int b = fromHex(hex_b);
toHex(a + b, hex_c);
}
Implementing the conversion functions left as an exercise.

Answer 2

Here is an example of how you can add hexadecimal numbers.
This simplistic code wraps around with 5-digit-overflow.
#include <iostream>
#include <exception>
#include <cstring>
using std::cout;
using std::endl;
struct BadDigit : public std::exception { };
int hexToDec( const char c ) {
if ( c >= '0' && c <= '9' ) {
return c -'0';
}
else if( c>= 'A' && c<='F') {
return c - 'A' + 10;
}
else if ( c >= 'a' && c<='f' ) {
return c - 'a' + 10;
}
else {
throw BadDigit();
}
}
char decToHex( int d ) {
if ( d >= 0 && d <= 9 ) {
return d + '0';
}
else if ( d >= 10 && d <= 15 ) {
return d - 10 + 'A';
}
else {
throw BadDigit();
}
}
void add( const char *a, const char *b, char *s )
{
int rem = 0;
for (int i = 4; i >= 0; i-- ) {
int ai = hexToDec(a[i]);
int bi = hexToDec(b[i]);
int ri = ai + bi + rem;
s[i] = decToHex( ri % 16 );
rem = ri / 16;
}
}
int main()
{
const char *hex1 = "effff";
const char *hex2 = "00001";
char result[7];
try {
add(hex1, hex2, result);
}
catch (std::exception& e) {
cout << "Exception: " << e.what() << endl;
return 1;
}
cout << hex1 << " + " << hex2 << " = " << result << endl;
return 0;
}