long-integer

I have made a little app in ios+firebase, now I am trying to connect android. In ios I save date as double (for example: -242528463.775282), but then I trying to retrieve same double in java it giving me another date. in IOS - 01.07.2009 in Java - 29.12.1969 double myDouble = date; long myLong = (long) (myDouble); System.out.println(myLong); Date itemDate = new Date(itemLong); String myDateStr = new SimpleDateFormat("dd-MM-yyyy").format(itemDate); editTextDate.setText(myDateStr); Is it possible to convert double to date without converting to long? Since your double represents the number of

Now I have been trying to learn Java Programming, I want to know why do we use things like Float , short , and int when we could be just be using Long and Double ? I don't understand that part. Great question, especially if you're coming from a language like JavaScript which does not make a distinction between types of numbers. Java is a bit more strict than those languages, and everything you write is first compiled to what is called byte code, which is sort of like assembly language, but it can only be read by the Java Virtual Machine (JVM). Because of this, you must specify exactly how many

I am using oracle verion 10. There is stored procedure in PL/SQL using varchar2 variable. The code is constantly appending the varchar2 variable. When the varchar2 variable size exceeds 32767, it cannot append any more value. Now I want to change the data type to long or clob(in order to accomodate more characters), it does not work. How to modify the code here to have the same appending functionality with clob or long? sample appending x:= x || 'mydata'; The long datatype is deprecated; if you can you should seriously consider migrating your long column to a clob . If you were working with a

I am referencing my java version JDK 1.8 but I am still getting error. What is wrong with this referencing (writing Java after 6 years)? or any other simpler way to do this? I did some search and these functions are available in later java versions. Eclipse is Oxygen The method valueOf(Long) is undefined for the type Long The method compareTo() is undefined for the type Long import java.util.Comparator; import java.lang.Long; public class MyComparator<Long> implements Comparator<Long>{ @Override public int compare(Long long1, Long long2) { //Long.compareTo() return Long.valueOf(long1)

How does a C compiler interpret the "L" which denotes a long integer literal, in light of automatic conversion? The following code, when run on a 32-bit platform (32-bit long, 64-bit long long), seems to cast the expression "(0xffffffffL)" into the 64-bit integer 4294967295, not 32-bit -1. Sample code: #include <stdio.h> int main(void) { long long x = 10; long long y = (0xffffffffL); long long z = (long)(0xffffffffL); printf("long long x == %lld\n", x); printf("long long y == %lld\n", y); printf("long long z == %lld\n", z); printf("0xffffffffL == %ld\n", 0xffffffffL); if (x > (long)

Why does this java code long a4 = 1L; long a3 = 1; long a2 = 100L * 1024 * 1024 * 1024; long a1 = 100 * 1024 * 1024 * 1024; System.out.println(a4); System.out.println(a3); System.out.println(a2); System.out.println(a1); when run, output 1 1 107374182400 0 instead of the expected 1 1 107374182400 107374182400 output? Richard Tingle 107374182400 is exactly 25 times the full range of an integer (2^32) , which means that if you try to fit it into an integer it will overflow . And because it would fit exactly 25 times it ends up precisely on 0 (this is a coincidence and other huge multiplications