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why the value for string shows in capital for the first print but then show small in second print? [duplicate]

This question already has answers here:
String replace method is not replacing characters
(5 answers)
Closed 3 months ago.
I have a string:
String c = "IceCream";
If I use toUpperCase() function then it returns the same string, but I want to get "ICECREAM".
Where is the problem?

The code
String c = "IceCream";
String upper = c.toUpperCase();
System.out.println(upper);
correctly prints "ICECREAM". However, the original string c isn't changed. Strings in Java are immutable so all operations on the string return a new copy.

Are you expecting the original variable, c, to have been changed by toUpperCase()? Strings are immutable; methods such as .toUpperCase() return new strings, leaving the original un-modified:
String c = "IceCream";
String d = c.toUpperCase();
System.out.println(c); // prints IceCream
System.out.println(d); // prints ICECREAM

The object can't be changed, because String is immutable. However, you can have the reference point to a new instance, which is all uppercase:
String c = "IceCream";
c = c.toUpperCase();

You're supposed to use it like this:
String c = "IceCream";
String upper_c = c.toUpperCase();

It could be a problem with your locale. Try:
String c = "IceCream";
return c.toUpperCase(Locale.ENGLISH);

Adding three different char in java as java variable and getting a number [duplicate]

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 4 years ago.
IntelliJ IDEA Capture
Why i am getting 152, I think it will give me an error.
Please explain it.
public class character {
public static void main(String[] args) {
char myCharValue1 = 'A';
char myCharValue2 = '2';
char myCharValue3 = '%';
System.out.println(myCharValue1 + myCharValue2 + myCharValue3);
}
}

That is because chars refer to a number, which in turn has an ASCII representation.
Looking at an ASCII table you can see that the chars A, 2 and % have following values respectivly: 65, 50 and 37.
Adding those numbers together, you'll end up with 152 which is what you got in your example.
To print out those chars you could use following:
System.out.printf("%s%s%s&n", myCharValue1 + myCharValue2 + myCharValue3);
Which will print A2% (and a newline)

The concatenation + is for String. What you're doing is adding the numeric values of your chars and printing them together.
If you start with "" and then use + as Patrick Parker shows in his comment, it will become concatenation instead of simple addition and you'll get the result you expect.

How does the compareTo method work for Strings? if a word comes after another word in the alphabet then it returns positive or negative? [duplicate]

This question already has answers here:
How does the compareTo() method, compare strings? [closed]
(2 answers)
Closed 4 years ago.
Let's say i write this:
String a = "Hello";
String b = "Goodbye";
int compare = a.compareToIgnoreCase(b);
System.out.println(compare);
What will the printout be?

It compares character by character for each string.
For example in this case
When a = "hello" and b = "Goodbye"
It checks first character of a with first character of b and computes the relative difference, if it's same then it checks the next character, else it computes the difference and returns it.
If string a is greater than string b it returns a positive difference else it returns a negative difference.

in this case the out are be 1, because H comes after of G

Why this java statement is generating this output? [duplicate]

This question already has answers here:
The concatenation of chars to form a string gives different results
(5 answers)
Closed 9 years ago.
I have a string str, let's say it's value is "hell".
The below statement returns "205hellhe" instead of "hehellhe"
return (str.charAt(0)) + (str.charAt(1))+str+(str.charAt(0)) + (str.charAt(1));
why (str.charAt(0)) + (str.charAt(1)) is returning 205 instead of "he" and why the same statement is returning "he" at the end?

Expressions are evaluated Left-To-Right.
Hence first two chars are evaluated and added, which results in a char (value = 205).
Next this char(=205) is added to a string, which results in String.
Hence the strange output.
Fix:
Use a StringBuilder instead
public static void main(String[] args) {
String str = "hell";
StringBuilder buff = new StringBuilder();
buff.append(str.charAt(0))
.append(str.charAt(1))
.append(str)
.append(str.charAt(0))
.append(str.charAt(1));
System.out.println(buff.toString());// prints 'hehellhe'
}

This is happening because the values of the ASCII codes of the characters are being added instead of being concatenated. In ASCII, h is represented as 104, and e as 101. The concatenation isn't working, probably because both operands in this case are chars/integers.

What's the meaning of char zero (0) in Java? [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Java string replace and the NUL (NULL, ASCII 0) character?
I'm doing some String algorithms in Java, and i noticed that wherever i include a char with the value of 0 (zero) it marks the end of the String. Like this:
String aString = "I'm a String";
char[] aStringArray = aString.toCharArray();
aStringArray[1] = 0;
System.out.println(new String(aStringArray)); //It outputs "I"
What's the reason/cause of this behaviour?

The character '\0' is the null character. It's a control character, and it does not terminate the string, that's not how strings work in Java (that's how they work in C, though.)

For some additional insight, add the following to you code:
System.out.println(new String(aStringArray).length());
for (Byte b : new String(aStringArray).getBytes()) {
System.out.print("["+b+"]");
}
Your rendering system (console or output window) is not displaying everything.

I'm wondering if you have posted your actual code. When using the String(byte[]) constructor the actual length of the string is dependent on the contents of the array.