I want to increment a number in a string in Java 8.
Like this:
String lifes = "3";
lifes--
But i know that way is impossible, does you know other way to do it?
Integer.parseInt("3") + 1
The above statement results 4 which means we are parsing the String value as Integer and then incrementing in here, hope it helps!
If the string only contains the number, I'd do something like this -
lifes = String.valueOf(Integer.parseInt(lifes) - 1);
Related
I have the following statement: binaryOutput = y % 2 + binaryOutput;
I want to the value of y % 2 to be added at the very beginning of binaryOutput. Is there a better way to do this operation than the way I did it? I feel like my method is a little redundant.
Edit: binaryOutput is a String object. y is an integer.
USe the insert method of StringBuilder:
_sb.insert(0, "Hello ");
Hello Im new to programming
and as title I was wonder how you can find last digit of any given number?
for example when entering in 5.51123123 it will display 3
All I know is I should use charAt
Should I use while loop?
thanks in advance
You would want to do something like this:
double number = 5.51123123;
String numString = number + "";
System.out.println(numString.charAt(numString.length()-1));
When you do the number + "", Java "coerces" the type of number from a double to a string and allows you to perform string functions on it.
The numString.length()-1 is because numString.length() returns the count of all the characters in the string BUT charAt() indexes into the string and its indexing begins at 0, so you need to do the -1 or you'll get a StringIndexOutOfBoundsException.
You can use below function simply and you can customize data types accordingly,
private int getLastDigit(double val){
String tmp = String.valueOf(val);
return Integer.parseInt(tmp.substring(tmp.length()-1));
}
You can't use charAt on a floating point variable (or any other data type other than Strings). (Unless you define a charAt method for your own classes...)
You can, however, convert that floating point number to a string and check the charAt(str.length()-1).
Convert your float variable to string and use charAt(strVar.length - 1).
double a=5.51123123;
String strVar=String.valueOf(a);
System.out.print(strVar.charAt(strVar.length -1);
Double doubleNo = 5.51123123;
String stringNo = doubleNo.toString();
System.out.println(stringNo.charAt(stringNo.length()-1));
This question already has answers here:
In Java, is the result of the addition of two chars an int or a char?
(8 answers)
Closed 9 years ago.
In a programming language called Java, I have the following line of code:
char u = 'U';
System.out.print(u + 'X');
This statement results in an output like this:
173
Instead of
UX
Am I missing something? Why isn't it outputing 'UX'? Thank you.
Because you are performing an addition of chars. In Java, when you add chars, they are first converted to integers. In your case, the ASCII code of U is 85 and the code for X is 88.
And you print the sum, which is 173.
If you want to concatenate the chars, you can do, for example:
System.out.print("" + u + 'X');
Now the + is not a char addition any more, it becomes a String concatenation (because the first parameter "" is a String).
You could also create the String from its characters:
char[] characters = {u, 'X'};
System.out.print(new String(characters));
In this language known as Java, the result of adding two chars, shorts, or bytes is an int.
Thus, the integer value of U (85) is added to the integer value of X (88) and you get an integer value of 173 (85+88).
To Fix:
You'll probably want to make u a string. A string plus a char will be a string, as will a string plus a string.
String u = "U"; // u is a string
System.out.print(u + 'X'); // string plus a char is a string
String u = "U";
System.out.print(u + "X");
instead of char type use String class or StringBuilder class
Another way to convert one of the characters to a string:
char u = 'U';
System.out.print(Character.toString(u) + 'X');
This way could be useful when your variable is of type char for a good reason and you can't easily redeclare it as a String.
That "language called Java" bit amused me.
A quick search for "java concatenate" (which I recommend you do now and every time you have a question) revealed that most good Java programmers hate using + for concatenation. Even if it isn't used numerically when you want string concatenation, like in your code, it is also slow.
It seems that a much better way is to create a StringBuffer object and then call its append method for each string you want to be concatenated to it. See here: http://www.ibm.com/developerworks/websphere/library/bestpractices/string_concatenation.html
i was reading and couldn't find quite the snippet. I am looking for a function that takes in a string and left pads zeros (0) until the entire string is 8 digits long. All the other snippets i find only lets the integer control how much to pad and not how much to pad until the entire string is x digits long. in java.
Example
BC238 => 000BC289
4 => 00000004
etc thanks.
If you're starting with a string that you know is <= 8 characters long, you can do something like this:
s = "00000000".substring(0, 8 - s.length()) + s;
Actually, this works as well:
s = "00000000".substring(s.length()) + s;
If you're not sure that s is at most 8 characters long, you need to test it before using either of the above (or use Math.min(8, s.length()) or be prepared to catch an IndexOutOfBoundsException).
If you're starting with an integer and want to convert it to hex with padding, you can do this:
String s = String.format("%08x", Integer.valueOf(val));
org.apache.commons.lang.StringUtils.leftPad(String str, int size, char padChar)
You can take a look here
How about this:
s = (s.length()) < 8 ? ("00000000".substring(s.length()) + s) : s;
or
s = "00000000".substring(Math.min(8, s.length())) + s;
I prefer using an existing library method though, such as a method from Apache Commons StringUtils, or String.format(...). The intent of your code is clearer if you use a library method, assuming it is has a sensible name.
The lazy way is to use something like: Right("00000000" + yourstring, 8) with simple implementations of the Right function available here: http://geekswithblogs.net/congsuco/archive/2005/07/07/45607.aspx
Let's say I have a time hh:mm (eg. 11:22) and I want to use a string tokenizer to split. However, after it's split I am able to get for example: 11 and next line 22. But how do I assign 11 to a variable name "hour" and another variable name "min"?
Also another question. How do I round up a number? Even if it's 2.1 I want it to round up to 3?
Have a look at Split a string using String.split()
Spmething like
String s[] = "11:22".split(":");;
String s1 = s[0];
String s2 = s[1];
And ceil for rounding up
Find ceiling value of a number using Math.ceil
Rounding a number up isn't too hard. First you need to determine whether it's a whole number or not, by comparing it cast as both an int and a double. If they don't match, the number is not whole, so you can add 1 to the int value to round it up.
// num is type double, but will work with floats too
if ((int)num != (double)num) {
int roundedNum = (int)num + 1;
}