Decimal to binary/hex using arrays - c++

Question

I'm having trouble with the hexadecimal part of my c++ program. When I use the switch for hexadecimal nothing returns. also for some reason my binary conversion has a leading 0 I cant seem to get rid of.
#include "stdafx.h"
#include <iostream>
#include <iomanip>
#include <array>
using namespace std;
void binary(int, int);
void hex(int, int);
int _tmain(int argc, _TCHAR* argv[])
{
int numb, base;
cout << "Enter a decimel number : ";
cin >> numb;
cout << "Enter a base you want number switched to: ";
cin >> base;
switch (base){
case 2: binary(numb, base); break;
case 8: break;
case 16: hex(numb, base); break;
default: break;
}
}
void binary(int numb, int base)
{
int bin[32] = { 0 };
int i = 0;
do
{
bin[i] = numb%base;
numb = numb / base;
i++;
} while (numb != 0);
cout << "\n";
while (i >= 0){
cout << bin[i];
i--;
}
cout << endl << endl;
}
void hex(int numb, int base){
int i;
int hex[10] = { 0 };
for (i=0; i > 10; i++){
hex[i] = numb%base;
numb = numb / base;
for (i; i > 0; i--)
{
if (hex[i] >= 10)
{
switch (hex[i]){
case 10: cout << "A"; break;
case 11: cout << "B"; break;
case 12: cout << "C"; break;
case 13: cout << "D"; break;
case 14: cout << "E"; break;
case 15: cout << "F"; break;
default: break;
}
}
cout << hex[i];
}
}
cout << endl;
}

Answer 1

binary
The problem is after the first loop, i is one greater than the last index. Just for example, say you enter 1: the do...while loop is entered, the digit 1 is put in array index 0, then i is incremented to 1.
Then, in the second loop, both indexes 1 and 0 are printed. You can solve this by decrementing i before entering this loop:
i--;
while (i >= 0){...}
You should be doing something like that anyway, because if you ended up using all 32 digits, you would try to access bin[32] and the program may crash or output gibberish.
hex
The first loop's condition is infinite:
for (i = 0; i >= 0; i++){...}
It should be the same as your condition in binary:
for (i = 0; numb != 0; i++){...}
But you are not done yet because I've noticed you also have a bug in your printing:
if (hex[i] >= 10)
{
switch (hex[i])
{
case 10:
cout << "A";
break;
...
}
}
cout << hex[i];
If hex[i] is greater than or equal to 10, it gets printed twice, once as a hex letter and once as a decimal number. To solve this you could, for example, use continue instead of break in your switch (to skip the second print), or use else:
if (hex[i] >= 10)
{
switch (hex[i])
{
case 10:
cout << "A";
break;
...
}
}
else
{
cout << hex[i];
}
You also need to make the same correction as in binary:
// decrementing i before entering the loop
// vvv
for (i--; i >= 0; i--){...}
Your revision is not correct, hex should not have a nested loop. It was fine before, just with the corrections I've pointed out.