C++ unix 32bit to Red Hat Linux running on an x86_64 chip (64-bit server) -

i working on project porting application in c++ 32 bit unix red hat linux running on x86_64 chip (64-bit server). facing issues related long data types.i wrote sample program print binary long , unsigned long value of 1. has 1 in middle:

ul = 0000000000000000000000000000000100000000000000000000000000000001
l = 0000000000000000000000000000000100000000000000000000000000000001

string binary(unsigned long i)    {    int n = sizeof(unsigned long);   string s;   n *= 8;   --n;   while (n >= 0)    {   unsigned long mask = (1 << n);    s += (mask & ? "1" : "0");    --n;   }    return s;   }    string binary(long i)   {   int n = sizeof( long);    string s;    n *= 8;   --n;   while (n >= 0)   {    long mask = (1 << n);    s += (mask & ? "1" : "0");   --n;    }    return s;   }    int main()   {   unsigned long ul = 1;   long l = 1;     cout << "sizeof ul = " << sizeof ul <<  ", ul = "  <<ul<<  endl;    cout << "sizeof l = "  << sizeof l  << ", l = " << l << endl;     cout << "ul = " << binary(ul) << endl;    cout << "l = "  << binary(l) << endl;     return 0;   }       

as can see not getting why there 1 in middle. causing run time issues in application. please let me know. getting 1 middle in red hat linux running on x86_64 chip (64-bit server) not in unix , cause of run time issues. idea please let me know.

the problem here:

unsigned long mask = (1 << n);   

apparently, unsigned long has 8 byte , normal int has 4 byte on system.

here have int value 1 (every literal value int if nothing else explicitely specified). you´re doing shifting on 4 byte int, , then result converted 8 byte save in mask. shifting more 31 bit in 4 byte, yet you´re doing until 63 bit.

to solve problem, have tell compiler 1 meant 8 byte unsigned long.
3 possibilites:

unsigned long mask = (1ul << n);   //or unsigned long mask = ((unsigned long)1 << n);   //or unsigned long mask = 1;   mask = mask << n;   

the other function (for signed type) has same problem.
first variant, use l there, without u.