CODEWITHSUNDEEP
c++negative-numberunsigned-long-long-int
user1739860
user1739860

Reputation:

Unsigned long returning negative in C++

below when any of the numA-D is multiplied numerous times the number suddenly becomes a negative. For example when numA is multiplied 3 times by 256 the number then becomes negative,but not if only multiplied by 256 twice. The purpose of this project is to change an ip address to a unsigned long, and then an unsigned long to an ip address.

using namespace std;
unsigned long ip2long (string ipv4)
{

// variable to return
unsigned long rtn;
string A,B,C,D;
string delimiter = ".";
size_t position;


/* must parse string by '.' character
and then assign (A,B,C,D) to the values
and return the unsigned long last, you don't have to make
the variables unsigned longs */


int locOfA = ipv4.find('.' );
int locOfB = ipv4.find('.', locOfA + 1);
int locOfC = ipv4.find('.', locOfB + 1);
int locOfD = ipv4.find('.', locOfC + 1);


A =  ipv4.substr(0,locOfA);
B = ipv4.substr(locOfA + 1, locOfB - locOfA - 1);
C = ipv4.substr(locOfB + 1, locOfC - locOfB -1 );
D = ipv4.substr(locOfC + 1, locOfD - locOfC -1);

int numA = atoi(A.c_str());
int numB = atoi(B.c_str());
int numC = atoi(C.c_str());
int numD = atoi(D.c_str());
cout << endl;
cout << numA << endl;
cout << numB << endl;
cout << numC << endl;
cout << numD << endl;


cout << endl;
// assigning a unsigned long to the sum of the algorithm
cout << (numA * 256 * 256) +  << endl;
cout << (256 * 256 * 256 * numB) << endl;
cout << (256 * numC) << endl;
cout << (256 * numD) << endl;

rtn = numD + (256 * numC) + (256 * 256 * numB) + (256 * 256 * 256 * numA);


return rtn;
}

Upvotes: 0

Views: 4060

Answers (3)

Ivan
Ivan

Reputation: 5250

All integer types are signed

  • LONG - Whitout unsigned is always signed –2,147,483,648 to 2,147,483,647
  • INT - Whitout unsigned is always signed –2,147,483,648 to 2,147,483,647
  • SHORT - Whitout unsigned is always signed –32,768 to 32,767
  • FLOAT - Whitout unsigned is always signed 3.4E +/- 38 (7 digits)
  • DOUBLE - Whitout unsigned is always signed 1.7E +/- 308 (15 digits)

The largest number that can be stored in unsigned was two times higher than signed.

Example:

long val = 2147483647; // Signed max value

unsigned long val = 4294967294;  // 2x bigger number can be stored

What if try you to store 4294967294 in signed?

signed long val = 4294967294;  // result will be -2

Like i say value in usigned can resive 2x bigger number. These errors occur in the case of exceeding

Upvotes: 0

idoo
idoo

Reputation: 330

int is 32 bit. it means its max value is 2^31 = 256*256*256*128 - 1. (and its lower value is -2^31)
if numA >=128 it will become negative.
or if numD + (256 * numC) + (256 * 256 * numB) + (256 * 256 * 256 * numA) > 2^31 = 2147483648
you will see negative number.

Upvotes: 0

Billy ONeal
Billy ONeal

Reputation: 106530

unsigned long is unsigned -- it can't be negative. But all of your numbers, and as a result all of your calculations, are ints, not unsigned longs. ints are allowed to be negative.

Put another way, this line

rtn = numD + (256 * numC) + (256 * 256 * numB) + (256 * 256 * 256 * numA);

Is the same as

rtn = static_cast<unsigned long>(numD + (256 * numC) + (256 * 256 * numB) + (256 * 256 * 256 * numA));

All the variables and constants are int, and they will not be automatically promoted to unsigned long.

Upvotes: 4

Related Questions