HTOI不正确的10位数字输出

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*10 ² + 2*10 ¹ + 3*10 ⁰

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;
}