long-integer

Wonder why. C# .Net 3.5 int a = 256 * 1024 * 1024; int b = 8; long c = b * a; Console.WriteLine(c);//<-- result is -2147483648 Where does this minus from? Where does this minus from? From the integer overflow. Note that your code is equivalent to: int a = 256 * 1024 * 1024; int b = 8; int tmp = b * a; long c = tmp; Console.WriteLine(c); I've separated out the multiplication from the assignment to the long variable to emphasize that they really are separate operations - the multiplication is performed using Int32 arithmetic, because both operands are Int32 - the fact that the result is assigned

Currently I'm brushing up on my Fortran95 knowledge (don't ask why)... I'm running in to a problem though. How does one handle large integers, eg. the size of: ~700000000000 INTEGER(KIND=3) cannot hold this number. If anyone is interested the compiler I have available is Silverfrost FTN95. I am using the integer to run through a larger set of data. Do you have any suggestions? The standard solution (since Fortran 95, so I assume your compiler supports it) is to use the SELECTED_INT_KIND intrinsic to probe for valid integer kinds (whose values are compiler dependent) and the HUGE intrinsic.

Wouldn't it have made more sense to make long 64-bit and reserve long long until 128-bit numbers become a reality? Yes, it does make sense, but Microsoft had their own reasons for defining "long" as 32-bits. As far as I know, of all the mainstream systems right now, Windows is the only OS where "long" is 32-bits. On Unix and Linux, it's 64-bit. All compilers for Windows will compile "long" to 32-bits on Windows to maintain compatibility with Microsoft. For this reason, I avoid using "int" and "long". Occasionally I'll use "int" for error codes and booleans (in C), but I never use them for any

I have a hashtable named table . The type value is long . I am getting values using .values() . Now I want to access these values. Collection val = table.values(); Iterator itr = val.iterator(); long a = (long)itr.next(); But when I try to get it, it gives me error because I can't convert from type object to long . How can I go around it? Try this: Long a = (Long)itr.next(); You end up with a Long object but with autoboxing you may use it almost like a primitive long. Another option is to use Generics: Iterator<Long> itr = val.iterator(); Long a = itr.next(); Number class can be used for to

I have an unsigned long long (or uint64_t ) value and want to convert it to a double . The double shall have the same bit pattern as the long value. This way I can set the bits of the double "by hand". unsigned long long bits = 1ULL; double result = /* some magic here */ bits; I am looking for a way to do this. The portable way to do this is with memcpy (you may also be able to conditionally do it with reinterpret_cast or a union, but those aren't certain to be portable because they violate the letter of the strict-alias rules): // First, static assert that the sizes are the same memcpy(

I read that JVM stores internally short, integer and long as 4 bytes. I read it from an article from the year 2000, so I don't know how true it is now. For the newer JVMs, is there any performance gain in using short over integer/long? And did that part of the implementation has changed since 2000? Thanks JonH long 64 –9,223,372,036,854,775,808 to 9 ,223,372,036,854,775,807 int 32 –2,147,483,648 to 2,147,483,647 short 16 –32,768 to 32,767 byte 8 –128 to 127 Use what you need, I would think shorts are rarely used due to the small range and it is in big-endian format. Any performance gain would