Htoi incorrect output at 10 digits - c++

Question

When I input
0x123456789
I get incorrect outputs, I can't figure out why. At first I thought it was a max possible int value problem, but I changed my variables to unsigned long and the problem was still there.
#include <iostream>
using namespace std;
long htoi(char s[]);
int main()
{
cout << "Enter Hex \n";
char hexstring[20];
cin >> hexstring;
cout << htoi(hexstring) << "\n";
}
//Converts string to hex
long htoi(char s[])
{
int charsize = 0;
while (s[charsize] != '\0')
{
charsize++;
}
int base = 1;
unsigned long total = 0;
unsigned long multiplier = 1;
for (int i = charsize; i >= 0; i--)
{
if (s[i] == '0' || s[i] == 'x' || s[i] == 'X' || s[i] == '\0')
{
continue;
}
if ( (s[i] >= '0') && (s[i] <= '9') )
{
total = total + ((s[i] - '0') * multiplier);
multiplier = multiplier * 16UL;
continue;
}
if ((s[i] >= 'A') && (s[i] <= 'F'))
{
total = total + ((s[i] - '7') * multiplier); //'7' equals 55 in decimal, while 'A' equals 65
multiplier = multiplier * 16UL;
continue;
}
if ((s[i] >= 'a') && (s[i] <= 'f'))
{
total = total + ((s[i] - 'W') * multiplier); //W equals 87 in decimal, while 'a' equals 97
multiplier = multiplier * 16UL;
continue;
}
}
return total;
}

Answer 1

long probably is 32 bits on your computer as well. Try long long.

Answer 2

You need more than 32 bits to store that number. Your long type could well be as small as 32 bits.
Use a std::uint64_t instead. This is always a 64 bit unsigned type. If your compiler doesn't support that, use a long long. That must be at least 64 bits.

Answer 3

The idea follows the polynomial nature of a number. 123 is the same as
1*102 + 2*101 + 3*100
In other words, I had to multiply the first digit by ten two times. I had to multiply 2 by ten one time. And I multiplied the last digit by one. Again, reading from left to right:
Multiply zero by ten and add the 1 → 0*10+1 = 1.
Multiply that by ten and add the 2 → 1*10+2 = 12.
Multiply that by ten and add the 3 → 12*10+3 = 123.
We will do the same thing:
#include <cctype>
#include <ciso646>
#include <iostream>
using namespace std;
unsigned long long hextodec( const std::string& s )
{
unsigned long long result = 0;
for (char c : s)
{
result *= 16;
if (isdigit( c )) result |= c - '0';
else result |= toupper( c ) - 'A' + 10;
}
return result;
}
int main( int argc, char** argv )
{
cout << hextodec( argv[1] ) << "\n";
}
You may notice that the function is more than three lines. I did that for clarity. C++ idioms can make that loop a single line:
for (char c : s)
result = (result << 4) | (isdigit( c ) ? (c - '0') : (toupper( c ) - 'A' + 10));
You can also do validation if you like. What I have presented is not the only way to do the digit-to-value conversion. There exist other methods that are just as good (and some that are better).
I do hope this helps.

Answer 4

I found out what was happening, when I inputted "1234567890" it would skip over the '0' so I had to modify the code. The other problem was that long was indeed 32-bits, so I changed it to uint64_t as suggested by #Bathsheba. Here's the final working code.
#include <iostream>
using namespace std;
uint64_t htoi(char s[]);
int main()
{
char hexstring[20];
cin >> hexstring;
cout << htoi(hexstring) << "\n";
}
//Converts string to hex
uint64_t htoi(char s[])
{
int charsize = 0;
while (s[charsize] != '\0')
{
charsize++;
}
int base = 1;
uint64_t total = 0;
uint64_t multiplier = 1;
for (int i = charsize; i >= 0; i--)
{
if (s[i] == 'x' || s[i] == 'X' || s[i] == '\0')
{
continue;
}
if ( (s[i] >= '0') && (s[i] <= '9') )
{
total = total + ((uint64_t)(s[i] - '0') * multiplier);
multiplier = multiplier * 16;
continue;
}
if ((s[i] >= 'A') && (s[i] <= 'F'))
{
total = total + ((uint64_t)(s[i] - '7') * multiplier); //'7' equals 55 in decimal, while 'A' equals 65
multiplier = multiplier * 16;
continue;
}
if ((s[i] >= 'a') && (s[i] <= 'f'))
{
total = total + ((uint64_t)(s[i] - 'W') * multiplier); //W equals 87 in decimal, while 'a' equals 97
multiplier = multiplier * 16;
continue;
}
}
return total;
}