Logical Error in C++ Hexadecimal Converter Code - linux

Question

I've been working on this Hexadecimal Converter and there seems to be a logical error somewhere in the program. I've run it on Ubuntu using the g++ tool and every time I run t program, it gives me a massive heap of garbage values. I can't figure out the source of the garbage values and neither can I find the source of the logical error. I'm a newbie at programming, so please help me figure out my mistake.
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
int bin[20],finhex[10],num,bc=0,i,j,k,l=0,r=10,n=1,binset=0,m=0;
int hex[16]= {0000,0001,0010,0011,0100,0101,0110,0111,1000,1001,1010,1011,1100,1101,1110,1111};
char hexalph='A';
cout<<"\nEnter your Number: ";
cin>>num;
while(num>0)
{
bin[bc]=num%2;
num=num/2;
bc++;
}
if(bc%4!=0)
bc++;
for(j=0;j<bc/4;j++)
for(i=0;i<4;i++)
{
binset=binset+(bin[m]*pow(10,i));
m++;
}
for(k=0;k<16;k++)
{
if(hex[k]==binset)
{
if(k<=9)
finhex[l]=k;
else
while(n>0)
{
if(k==r)
{
finhex[l]=hexalph;
break;
}
else
{
hexalph++;
r++;
}
}
l++;
r=10;
binset=0;
hexalph='A';
break;
}
}
while(l>=0)
{
cout<<"\n"<<finhex[l];
l--;
}
return 0;
}

Answer 1

int hex[16]= {0000,0001,0010,0011,0100,0101,0110,0111,1000,1001,1010,1011,1100,1101,1110,1111};
Allow me to translate those values into decimal for you:
int hex[16] = {0, 1, 8, 9, 64, 65, 72, 73, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111};
If you want them to be considered binary literals then you need to either specify them as such or put them in some other form that the compiler understands:
int hex[16] = {0b0000, 0b0001, 0b0010, 0b0011, 0b0100, 0b0101, 0b0110, 0b0111, 0b1000, 0b1001, 0b1010, 0b1011, 0b1100, 0b1101, 0b1110, 0b1111};
int hex[16] = {0x0, 0x1, 0x2, 0x3, 0x4, 0x5, 0x6, 0x7, 0x8, 0x9, 0xa, 0xb, 0xc, 0xd, 0xe, 0xf};

Answer 2

While Ignacio Vazquez-Abrams rightly hinted at the fact that some of your initializers of hex are (due to the prefix 0) octal constants, he overlooked that you chose the unusual, but possible way of representing binary literals as decimal constants with only the digits 0 and 1. Thus, you only have to remove the prefix 0 from all constants greater than 7:
int hex[16] = {0000,0001,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111};
Then, you stored the characters 'A' etc. in int finhex[] and did output them with cout<<"\n"<<finhex[l] - but this way not A is printed, but rather its character code value, e. g. in ASCII 65. In order to really output the character A etc., we could change the finhex array element type to char:
char bin[20],finhex[10]; int num,bc=0,i,j,k,l=0,r=10,n=1,binset=0,m=0;
- but consequently we also have to store the digits 0 to 9 as their character code values:
if (k<=9)
finhex[l]='0'+k;
Furthermore, with the lines
if(bc%4!=0)
bc++;
you rightly pondered on the need to have a multiple of 4 bits for the conversion, but you overlooked that more than one bit could be missing, and also that the additional elements of bin[] are uninitialized, so change to:
while (bc%4!=0) bin[bc++] = 0;
Besides, you omitted block braces around the (appropriately indented) two inner for loops; since C++ is not Python, the indentation has no significance and without surrounding braces only the first of the indented for loops is nested into the outer for loop.
The final while loop should be outdented and go outside the big for loop. There's also an indexing error in it, as the finhex array is indexed with an l which is by one to high; you could change this to:
while (l--) cout<<finhex[l];
cout<<"\n";