long-integer

I've spent hours scouring the internet for a solution to this problem and can't find anything. I have been trying to write unformatted output to a CSV output file with multiple very long lines of varying length and multiple data types. I'm trying to first write a long header that indicates the variables that will be written below, separated by commas. Then on the lines below that, I am writing the values specified in the header. However, with sequential access, the long output lines are broken into multiple shorter lines, which is not what I was hoping for. I tried controlling the line length

For example, A=10^17, B=10^17, C=10^18. The product A*B exceeds the limit of long long int. Also, writing ((A%C)*(B%C))%C doesn't help. Assuming you want to stay within 64-bit integer operations, you can use binary long division, which boils down to a bunch of adds and multiply by two operations. This means you also need overflow-proof versions of those operators, but those are relatively simple. Here is some Java code that assumes A and B are already positive and less than M. If not, it's easy to make them so beforehand. // assumes a and b are already less than m public static long addMod

The Python tutorial book I'm using is slightly outdated, but I've decided to continue using it with the latest version of Python to practice debugging. Sometimes there are a few things in the book's code that I learn have changed in the updated Python, and I'm not sure if this is one of them. While fixing a program so that it can print longer factorial values, it uses a long int to solve the problem. The original code is as follows: #factorial.py # Program to compute the factorial of a number # Illustrates for loop with an accumulator def main(): n = input("Please enter a whole number: ") fact

long l2 = 32; When I use the above statement, I don't get an error (I did not used l at the end), but when I use the below statement, I get this error: The literal 3244444444 of type int is out of range long l2 = 3244444444; If I use long l2 = 3244444444l; , then there's no error. What is the reason for this? Using l is not mandatory for long variables. 3244444444 is interpreted as a literal integer but can't fit in a 32-bit int variable. It needs to be a literal long value , so it needs an l or L at the end: long l2 = 3244444444l; // or 3244444444L More info: Primitive Data Types ,

I got a very strange problem when I'm trying to compare 2 Long variables, they always show false and I can be sure they have the same number value by debugging in Eclipse: if (user.getId() == admin.getId()) { return true; // Always enter here } else { return false; } Both of above 2 return values are object-type Long, which confused me. And to verify that I wrote a main method like this: Long id1 = 123L; Long id2 = 123L; System.out.println(id1 == id2); It prints true. So can somebody give me ideas?. I've been working in Java Development for 3 years but cannot explain this case. BlackJoker ==

How can I get a long number bigger than Long.MAX_VALUE? I want this method to return true : boolean isBiggerThanMaxLong(long val) { return (val > Long.MAX_VALUE); } That method can't return true . That's the point of Long.MAX_VALUE . It would be really confusing if its name were... false. Then it should be just called Long.SOME_FAIRLY_LARGE_VALUE and have literally zero reasonable uses. Just use Android's isUserAGoat , or you may roll your own function that always returns false . Note that a long in memory takes a fixed number of bytes. From Oracle : long: The long data type is a 64-bit signed

I've spent the past two hours debugging what seems extremely unlikely. I've stripped the method of a secondary Android Activity to exactly this: public void onClick(View v) { String str = "25"; long my_long = Long.getLong(str); } // onClick (v) And yeah, I get a crash with the good ol' NullPointerException: 09-11 02:02:50.444: ERROR/AndroidRuntime(1588): Uncaught handler: thread main exiting due to uncaught exception 09-11 02:02:50.464: ERROR/AndroidRuntime(1588): java.lang.NullPointerException It looks like (from other tests) that Long.getLong(str) returns NULL, which is driving me bonkers.

Does Lua make use of 64-bit integers? How do I use it? Compile it yourself. Lua uses double-precision floating point numbers by default. However, this can be changed in the source ( luaconf.h , look for LUA_NUMBER ). require "bit" -- Lua unsigned 64bit emulated bitwises -- Slow. But it works. function i64(v) local o = {}; o.l = v; o.h = 0; return o; end -- constructor +assign 32-bit value function i64_ax(h,l) local o = {}; o.l = l; o.h = h; return o; end -- +assign 64-bit v.as 2 regs function i64u(x) return ( ( (bit.rshift(x,1) * 2) + bit.band(x,1) ) % (0xFFFFFFFF+1)); end -- keeps [1+0.

How should I perform conversion from IPv6 to long and vice versa? So far I have: public static long IPToLong(String addr) { String[] addrArray = addr.split("\\."); long num = 0; for (int i = 0; i < addrArray.length; i++) { int power = 3 - i; num += ((Integer.parseInt(addrArray[i], 16) % 256 * Math.pow(256, power))); } return num; } public static String longToIP(long ip) { return ((ip >> 24) & 0xFF) + "." + ((ip >> 16) & 0xFF) + "." + ((ip >> 8) & 0xFF) + "." + (ip & 0xFF); } Is it correct solution or I missed something? (It would be perfect if the solution worked for both ipv4 and ipv6) Andrei