This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How to convert number to words in java
I am trying to set the long value in the textfield the modification of my code is as below error what it is showing is required long found int
private void jTextField2MouseClicked(java.awt.event.MouseEvent evt) {
if(evt.getSource()==jTextField2){
long jml = Long.parseLong(jTextField3.getText());
jTextField1.setText(numberToWord(jml));
}
}
The NumberFormatException is Thrown to indicate that the application has attempted to convert a string to one of the numeric types, but that the string does not have the appropriate format. So, the exception could be because the maximum value of a java int is 2,147,483,647 (which is a 10-digit integer).
The error is occuring in the below line, And looks like you are using Integer.parseInt() method in the below line,
at myproj.Certificates.jTextField2MouseClicked(Certificates.java:268)
When the value that is being passed to the method Integer.parseInt() is greater than Integer.MAX_VALUE (2,147,483,647), you will receive the java.lang.NumberFormatException.
And since 10000000000 is greater than 2,147,483,647 you are getting the NumberFormatException.
Related
This question already has answers here:
What is the ellipsis (...) for in this method signature?
(5 answers)
What do 3 dots next to a parameter type mean in Java?
(9 answers)
Closed 7 years ago.
I came across a method definition:
void doType(int... keyCodes) {
doType(keyCodes, 0, keyCodes.length);
}
void doType(int[] keyCodes, int offset, int length) {
if (length == 0) {
return;
}
robot.keyPress(keyCodes[offset]);
doType(keyCodes, offset + 1, length - 1);
robot.keyRelease(keyCodes[offset]);
}
The "int..." seems to indicate an indeterminate number of integer parameters but the variable is used as an array inside the function. Can someone please explain?
As you already stated correctly this java notation is to make the method accept a variable amount of (in this case) int parameter.
To handle this variable amount of variables you can access it like an array.
This functionality is introduced in java 5.
See also here:
https://docs.oracle.com/javase/1.5.0/docs/guide/language/varargs.html
You are right in deducing that the ellipses indicate that this method is variadic.
When you have a variable number of potential arguments, you need some way to iterate over them - otherwise they aren't very useful in practice. Java and C# happen to have the array indexing syntax. C and C++ happen to have the va_start, va_arg and va_end macros. Different languages may have something else.
The reason why Java has the array syntax specifically is probably because it happens to match the way they are actually implemented: as a simple array parameter replaced at compile time.
This question already has answers here:
Extract Integer Part in String
(9 answers)
Closed 7 years ago.
Is there a way to get a Integer variable from a String object, something like:
String string = "Render.SCREEN_WIDTH_TILES";
// SCREEN_WIDTH_TILES is a Render Integer value referenced in other class
I want the string variable to hold the reference to the specified int.
What I want from that string value is to "transform it" into a int value,
Is it possible to do this?
I can't find a way to handle an integer value as a variable in a string.
You seem to be misunderstanding how Integer.parseInt(s) works. The documentation clearly states:
Parses the string argument as a signed decimal integer. The characters in the string must all be decimal digits, except that the first character may be an ASCII minus sign - (\u002D) to indicate a negative value or an ASCII plus sign + (\u002B) to indicate a positive value.
The parameter to Integer.parseInt(s) must be a string that contains a number. Such as:
Integer.parseInt("12345")
Integer.parseInt("-45")
But not:
Integer.parseInt("Hello world")
Integer.parseInt("this will not work 4 you")
And certainly not: Integer.parseInt("Render.SCREEN_WIDTH_TILES - 1");
This question already has answers here:
Reading a file larger than 2GB into memory in Java
(4 answers)
Closed 7 years ago.
I have a very long String upto 10^15 characters. I need to see what character is at a particular index but charAt() method takes integer parameter. How can I find character at a particular index (>10^9) ?
Edit: I figured out I can't store String of order 10^15 but I am wondering is there a particular class in java for storing very large strings?
In the String class, the characters are stored in an array (private final char value[]), and Java arrays are indexed with an int.
This explains why the charAt() method takes an int as a parameter instead of a long, and why you won't be able to store any string with more than 2^31 characters anyway.
This question already has answers here:
The literal xyz of type int is out of range
(5 answers)
Closed 9 years ago.
I have to use quite a big number - 600851475143; as we all know, I have to use long data type but when I try to initialize like: long number = 600851475143 I get an error:
The literal 600851475143 of type int is out of range.
It seems that I don't know how to use long data type correctly.
long number = 600851475143L
Use "L" to make it as long type
Use "L" to make it as long type. By default all integer type variable(byte,int,long) is "int"
long num=600851475143L;
or
long num=600851475143l; // small 'L'
Use
long number = 600851475143L;
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Integer with leading zeroes
The program I am coding requires me to label an item with an inventory number of 012345 and store it in a int variable.
This is a stripped down example of what I am doing:
int test = 012345;
System.out.println(test);
this prints as:
5349
How do I get it to print out as 012345 rather than 5349?
EDIT: I am entering this into the parameter of a constructor for a custom class i am initializing. Then I use a method to return what the current number is, then print it to the terminal window.
You get a wrong number because when you prepend zero to an integer literal, Java interprets the number as an octal (i.e. base-8) constant. If you want to add a leading zero, use
int test = 12345;
System.out.println("0"+test);
You can also use the formated output functionality with the %06d specifier, like this:
System.out.format("%06d", num);
6 means "use six digits"; '0' means "pad with zeros if necessary".
As already said, int value with leading zero is considered as octal value. If you don't need to have test as int, why not make it string? Like
String test= new String("012345");
And if you want to use int for test, you can do not prepend 0, rather just use the number and prepend 0 at the time of printing.
In case if you're wondering how will you find how many leading zero are to be prepended, you may do like this
int lengthOfItemID=6;
int test=12345;
String test1=new String("000000"+test);
System.out.println(test1.substring(test1.length()-lengthOfItemID));
Pardon syntax mistakes, been years I last worked with java.
You can get the right result by using Integer.parseInt. That will make your string into a decimal string. (found here). The JAVA API here states that it takes a string and returns a signed decimal.