I'm new to java language I confused about different between these 2 methods calls (I don't know exactly it is method call or not)
but if i need to call method to calculate something I call this
salary obj=new salary();
obj.bonus(45000);
but when we call String length method we call this as
String name="Java lovely";
name.length();
What are the different of these 2 methods .Please help me to overcome this problem.
Thank you
String name="Java lovely";
You might confused with the string literal.
String is special in java
You will clear if I wrote
String s = new String("Java Lovely");
s.lenght();
String is a class. String class represents character strings. All string literals in Java programs, such as "abc", are implemented as instances of this class.
Strings are constant; their values cannot be changed after they are created. String buffers support mutable strings. Because String objects are immutable they can be shared. For example:
String str = "Java lovely";
is equivalent to:
char data[] = {'J', 'a', 'v', 'a', ' ', 'l','o', 'v', 'e','l', 'y'};
String str = new String(data);
length is a method of String Class.
in upper part that is
salary obj=new salary();
obj.bonus(45000);
in this salary is a class and you created object of this 'obj' and bonus is a method that you defined in salary class and you are calling that method
and other is
String name="Java lovely";
name.length();
is here same
String name=new String("Java lovely");
and you are calling length method on String object 'name'.
both are same but one difference is String is constant(immutable) mean once you created it, you can't change it but your salary class is not constant,you can change it.
To understand java better u can read this book. Head first Java
It is exactly same. Both obj and name are instance of their class. BUt String is exceptional class in java.. It can be create like above without new keyword.
it's the same, as String is a ready included java class.
you can make :
String name = new String("dsdsd");
it seems like
String name = new String();
name = "dsdsd";
name.length();
What do you men by difference?
That the first call has something between the () ? And name.length not? That is because the method bonus is an method with parameters. It expects an value, at this case an integer or int to do something with this. but Name.length doesnt mean any values, it just returns the saved length of the string.
Or where else you see a difference?
obj.bonus(number); and
name.length ();
is actually the same...
Sry if this wasnt your question.
Edit: or do you mean the constructor? This is because string is special in java, actually new String("lalala"); is right.
salary obj=new salary(); // You are initializing salary class
obj.bonus(45000); // Your salary class contains bonus method and it takes
int argument
So you are call bonus method and parse 45000 to that.
String name="Java lovely"; // You are define a String
name.length(); // this will return the length of name String, no need input
argument and return int length
Related
I wanted to create a char array of the alphabet. I looked at this post:
Better way to generate array of all letters in the alphabet
which said this:
char[] alphabet = "abcdefghijklmnopqrstuvwxyz".toCharArray();
So in my code I have:
public class Alphabet {
private char[] letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".toCharArray();
public String availableLetters(){
return letters.toString();
}
}
When I call the function availableLetters() from main() and printit to the console, it outputs this garbage:
[C#15db9742
What am I doing wrong?
The array is correct, the problem is that you are not printing it correctly.
If you print your array one character at a time, you would get a correct result:
for (char c : letters) {
System.out.print("'" + c + "' ");
}
demo
Unfortunately, Java standard class library does not provide a meaningful override of toString() for arrays, causing a lot of trouble for programmers who are new to the language.
If you want to print it in array form, then use:
System.out.println(Arrays.toString(letters));
BTW: The [C#15db9742 is not really garbage. It's what gets printed out when a class does not override the toString() method.
From Object.toString():
Returns a string representation of the object. In general, the
toString method returns a string that "textually represents" this
object. The result should be a concise but informative representation
that is easy for a person to read. It is recommended that all
subclasses override this method. The toString method for class Object
returns a string consisting of the name of the class of which the
object is an instance, the at-sign character `#', and the unsigned
hexadecimal representation of the hash code of the object. In other
words, this method returns a string equal to the value of:
getClass().getName() + '#' + Integer.toHexString(hashCode())
You can pass the char array to the String constructor or the static method String.valueOf() and return that instead.
I had used the following code to find the length
String str=strLenData.toString();
int ipLen= str.length();
return ipLen;
ipLen would return 11 every time. whatever be the actual value of strLenData. when I call toString() function, value of str: "[C#40523f80". Now I have to use char[] and i need to know the end (or length) of char[].
How do I do it?
If I understand correctly the strLenData variable is a char[]? In that case, you can just do
return strLenData.length;.
I don't think strLenData is charSequence or something. its just a regular object and may be it does't have toString() method. so toString() method of Object class gets called which just returns the HashCode value of The Objects Reference.
so make sure strLenData.toString() can be used.
ipLen is returning 11 always because you are converting an char array to String equivalent by using toString() method.
if you want to create String by an char array use.
String str = new String(strLenData);
Or if you want just want to get length of your char array use answer given by Steven.
use strLenData.length;
How do String objects work in Java? How does term "immutable" exactly apply to string objects? Why don't we see modified string after passing through some method, though we operate on original string object value?
a String has a private final char[] . when a new String object is created, the array is also created and filled. it cannot be later accessed [from outside] or modified [actually it can be done with reflection, but we'll leave this aside].
it is "immutable" because once a string is created, its value cannot be changed, a "cow" string will always have the value "cow".
We don't see modified string because it is immutable, the same object will never be changed, no matter what you do with it [besides modifying it with reflection]. so "cow" + " horse" will create a new String object, and NOT modify the last object.
if you define:
void foo(String arg) {
arg= arg + " horse";
}
and you call:
String str = "cow";
foo(str);
the str where the call is is not modified [since it is the original reference to the same object!] when you changed arg, you simply changed it to reference another String object, and did NOT change the actual original object. so str, will be the same object, which was not changed, still containing "cow"
if you still want to change a String object, you can do it with reflection. However, it is unadvised and can have some serious side-affects:
String str = "cow";
try {
Field value = str.getClass().getDeclaredField("value");
Field count = str.getClass().getDeclaredField("count");
Field hash = str.getClass().getDeclaredField("hash");
Field offset = str.getClass().getDeclaredField("offset");
value.setAccessible(true);
count.setAccessible(true);
hash.setAccessible(true);
offset.setAccessible(true);
char[] newVal = { 'c','o','w',' ','h','o','r','s','e' };
value.set(str,newVal);
count.set(str,newVal.length);
hash.set(str,0);
offset.set(str,0);
} catch (NoSuchFieldException e) {
} catch (IllegalAccessException e) {}
System.out.println(str);
}
From the tutorial:
The String class is immutable, so that once it is created a String object cannot be changed. The String class has a number of methods, some of which will be discussed below, that appear to modify strings. Since strings are immutable, what these methods really do is create and return a new string that contains the result of the operation.
Strings in Java are immutable (state cannot be modified once created). This offers opportunities for optimization. One example is string interning, where string literals are maintained in a string pool and new String objects are only created if the particular string literal doesn't already exist in the pool. If the string literal already exists, a reference is returned. This can only be accomplished because strings are immutable, so you don't have to worry that some object holding a reference will change it.
Methods that appear to modify a string actually return a new instance. One example is string concatenation:
String s = "";
for( int i = 0; i < 5; i++ ){
s = s + "hi";
}
What actually happens internally (the compiler changes it):
String s = "";
for( int i = 0; i < 5; i++ ){
StringBuffer sb = new StringBuffer();
sb.append(s);
sb.append("hi");
s = sb.toString();
}
You can clearly see that new instances are created by the toString method (note that this can be made more efficient by directly using StringBuffers). StringBuffers are mutable, unlike Strings.
Every object has state. The state of a String object is the array of characters that make up the String, for example, the String "foo" contains the array ['f', 'o', 'o']. Because a String is immutable, this array can never be changed in any way, shape, or form.
Every method in every class that wants to change a String must instead return a new String that represents the altered state of the old String. That is, if you try to reverse "foo" you will get a new String object with internal state ['o', 'o', 'f'].
I think this link will help you to understand how Java String really works
Now consider the following code -
String s = "ABC";
s.toLowerCase();
The method toLowerCase() will not change the data "ABC" that s contains. Instead, a new String object is instantiated and given the data "abc" during its construction. A reference to this String object is returned by the toLowerCase() method. To make the String s contain the data "abc", a different approach is needed.
Again consider the following - s = s.toLowerCase();
Now the String s references a new String object that contains "abc". There is nothing in the syntax of the declaration of the class String that enforces it as immutable; rather, none of the String class's methods ever affect the data that a String object contains, thus making it immutable.
I don't really understood your third question. May be providing a chunk of code and telling your problem is a better option. Hope this helps.
You can also look into this blogpost for more understanding
[code samples are taken from the wiki. you can also look in there for more information]
I know that at compile time when a String is created, that String will be THE string used by any objects of that particular signature.
String s = "foo"; <--Any other identical strings will simply be references to this object.
Does this hold for strings created during methods at runtime? I have some code where an object holds a piece of string data. The original code is something like
for(datum :data){
String a = datum.getD(); //getD is not doing anything but returning a field
StringBuffer toAppend = new StringBuffer(a).append(stuff).toString();
someData = someObject.getMethod(a);
//do stuff
}
Since the String was already created in data, it seems better to just call datum.getD() instead of creating a string on every iteration of the loop.
Unless there's something I'm missing?
String instances are shared when they are the result of a compile-time constant expression. As a result, in the example below a and c will point to the same instance, but b will be a different instance, even though they all represent the same value:
String a = "hello";
String b = hell() + o();
String c = "hell" + "o";
public String hell() {
return "hell";
}
public String o() {
return "o";
}
You can explicitly intern the String however:
String b = (hell() + o()).intern();
In which case they'll all point to the same object.
The line
String a = datum.getD();
means, assign the result of evaluating datum.getD() to the reference a . It doesn't create a new String.
You are correct that strings are immutable so all references to the same string value use the same object.
As far as being static, I do not think Strings are static in the way you describe. The Class class is like that, but I think it is the only object that does that.
I think it would be better to just call the datum.getD() since there is nothing that pulling it out into its own sting object gains for you.
If you do use the datum.getD() several times in the loop, then it might make sense to pull the value into a String object, because the cost of creating a string object once might be less than the cost of calling the getD() function multiple times.
Here's where I wish Java's String class had a replaceLast method, bu it doesn't and I'm getting the wrong results with my code.
I'm writing a program that searches a data structure for any items that match a string prefix. However, since I'm using an Iterator, the last item returned by the iter.next() call doesn't match the pattern, so I want to change the search string so that the last character of the query is increased by one letter. My testing code is returning [C#b82368 with this code and An as titleSearch:
public String changeLastCharacter(String titleSearch) {
char[] temp= titleSearch.toCharArray();
char lastLetter= temp[temp.length-1];
lastLetter++;
temp[temp.length-1]= lastLetter;
String newTitleSearch= temp.toString();
return newTitleSearch;
}
First, what is the cause of the output from this code?
Second, is there a better way to execute my solution?
You want:
newTitleSearch = new String(temp);
The toString method is not overridden for arrays; it's the usual Object.toString, intended for debugging. The above actually creates a string of the characters. An alternative is:
int len = titleSearch.length();
String allButLast = titleSearch.substring(0, len - 1);
newTitleSearch = allButLast + new Character(titleSearch.charAt(len - 1) + 1);
Whenever you see unexpected output like ....#<hex-digits>, the chances are that you are accidentally using toString() on some object whose class inherits the default implementation from Object.
The default toString() method returns a String whose value consists of the type name for the object combined with the object's "identity hash code" as hex digits. In your case the [C part is the type name for a char[] object. The '[' means "array of" and the 'C' means the char primitive type.
The rules for forming the type names used in the default toString() method are fully documented in the javadocs for java.lang.Class.getName().
Your problem is temp.toString(). Try String newTitleSearch = new String(temp); instead.
I figured this out by System.out.println(temp[0]+","+temp[1]); after temp[1] add been assigned to the incremented value. You could do this even easier by using your IDE's debugger.
Since the array was being assigned properly, the problem had to be in the toString().