i've a List<Polygon> polygons, where Polygon represents the geojson concept of polygon. In the class Polygon i defined a method toGeojson() that returns a string containing the geojson representation. I'd like to print all the list in a compact way instead of doing this:
String result = '';
for(Polygon p: polygons)
result += p.toGeojson();
I could do result = p.toString() but i cannot use toString() method because i use it for an other thing. Is there a way to call toGeojson() on a List just as you'd do with toString()?
Not sure if that answers your question, but you can use Stream api for that thing.
String result = polygons.stream()
.map(Polygon::toGeojson)
.collect(Collectors.joining(","));
There is no direct way to override behaviour of List.toString().
updated
There is Collectors#joining(CharSequence delimiter, CharSequence prefix, CharSequence suffix) method which accepts suffix and prefix. Using this method we can make our output look exactly like List.toSting with square brackets.
String result = polygons.stream()
.map(Polygon::toGeojson)
.collect(Collectors.joining(",", "[", "]")); // ["x","y"]
I am not sure I understand what you want, but I guess you are looking for a way to print the geoJson representation of each Polygon contained in your List. In that case I don't see a better way than a loop, but avoid String concatenation inside loops. Use StringBuilder instead which has much better performance.
StringBuilder result = new StringBuilder();
for (Polygon p: polygons) {
result.append(p.toGeojson());
}
Your solution is the best, i think... In Java there is no faster solution and the Array.toString method works the same way.
I'm a C++ guy learning Java. I'm reading Effective Java and something confused me. It says never to write code like this:
String s = new String("silly");
Because it creates unnecessary String objects. But instead it should be written like this:
String s = "No longer silly";
Ok fine so far...However, given this class:
public final class CaseInsensitiveString {
private String s;
public CaseInsensitiveString(String s) {
if (s == null) {
throw new NullPointerException();
}
this.s = s;
}
:
:
}
CaseInsensitiveString cis = new CaseInsensitiveString("Polish");
String s = "polish";
Why is the first statement ok? Shouldn't it be
CaseInsensitiveString cis = "Polish";
How do I make CaseInsensitiveString behave like String so the above statement is OK (with and without extending String)? What is it about String that makes it OK to just be able to pass it a literal like that? From my understanding there is no "copy constructor" concept in Java?
String is a special built-in class of the language. It is for the String class only in which you should avoid saying
String s = new String("Polish");
Because the literal "Polish" is already of type String, and you're creating an extra unnecessary object. For any other class, saying
CaseInsensitiveString cis = new CaseInsensitiveString("Polish");
is the correct (and only, in this case) thing to do.
I believe the main benefit of using the literal form (ie, "foo" rather than new String("foo")) is that all String literals are 'interned' by the VM. In other words it is added to a pool such that any other code that creates the same string will use the pooled String rather than creating a new instance.
To illustrate, the following code will print true for the first line, but false for the second:
System.out.println("foo" == "foo");
System.out.println(new String("bar") == new String("bar"));
Strings are treated a bit specially in java, they're immutable so it's safe for them to be handled by reference counting.
If you write
String s = "Polish";
String t = "Polish";
then s and t actually refer to the same object, and s==t will return true, for "==" for objects read "is the same object" (or can, anyway, I"m not sure if this is part of the actual language spec or simply a detail of the compiler implementation-so maybe it's not safe to rely on this) .
If you write
String s = new String("Polish");
String t = new String("Polish");
then s!=t (because you've explicitly created a new string) although s.equals(t) will return true (because string adds this behavior to equals).
The thing you want to write,
CaseInsensitiveString cis = "Polish";
can't work because you're thinking that the quotations are some sort of short-circuit constructor for your object, when in fact this only works for plain old java.lang.Strings.
String s1="foo";
literal will go in pool and s1 will refer.
String s2="foo";
this time it will check "foo" literal is already available in StringPool or not as now it exist so s2 will refer the same literal.
String s3=new String("foo");
"foo" literal will be created in StringPool first then through string arg constructor String Object will be created i.e "foo" in the heap due to object creation through new operator then s3 will refer it.
String s4=new String("foo");
same as s3
so System.out.println(s1==s2);// **true** due to literal comparison.
and System.out.println(s3==s4);// **false** due to object
comparison(s3 and s4 is created at different places in heap)
Strings are special in Java - they're immutable, and string constants are automatically turned into String objects.
There's no way for your SomeStringClass cis = "value" example to apply to any other class.
Nor can you extend String, because it's declared as final, meaning no sub-classing is allowed.
Java strings are interesting. It looks like the responses have covered some of the interesting points. Here are my two cents.
strings are immutable (you can never change them)
String x = "x";
x = "Y";
The first line will create a variable x which will contain the string value "x". The JVM will look in its pool of string values and see if "x" exists, if it does, it will point the variable x to it, if it does not exist, it will create it and then do the assignment
The second line will remove the reference to "x" and see if "Y" exists in the pool of string values. If it does exist, it will assign it, if it does not, it will create it first then assignment. As the string values are used or not, the memory space in the pool of string values will be reclaimed.
string comparisons are contingent on what you are comparing
String a1 = new String("A");
String a2 = new String("A");
a1 does not equal a2
a1 and a2 are object references
When string is explicitly declared, new instances are created and their references will not be the same.
I think you're on the wrong path with trying to use the caseinsensitive class. Leave the strings alone. What you really care about is how you display or compare the values. Use another class to format the string or to make comparisons.
i.e.
TextUtility.compare(string 1, string 2)
TextUtility.compareIgnoreCase(string 1, string 2)
TextUtility.camelHump(string 1)
Since you are making up the class, you can make the compares do what you want - compare the text values.
The best way to answer your question would be to make you familiar with the "String constant pool". In java string objects are immutable (i.e their values cannot be changed once they are initialized), so when editing a string object you end up creating a new edited string object wherease the old object just floats around in a special memory ares called the "string constant pool". creating a new string object by
String s = "Hello";
will only create a string object in the pool and the reference s will refer to it, but by using
String s = new String("Hello");
you create two string objects: one in the pool and the other in the heap. the reference will refer to the object in the heap.
You can't. Things in double-quotes in Java are specially recognised by the compiler as Strings, and unfortunately you can't override this (or extend java.lang.String - it's declared final).
- How do i make CaseInsensitiveString behave like String so the above statement is ok (with and w/out extending String)? What is it about String that makes it ok to just be able to pass it a literal like that? From my understanding there is no "copy constructor" concept in Java right?
Enough has been said from the first point. "Polish" is an string literal and cannot be assigned to the CaseInsentiviveString class.
Now about the second point
Although you can't create new literals, you can follow the first item of that book for a "similar" approach so the following statements are true:
// Lets test the insensitiveness
CaseInsensitiveString cis5 = CaseInsensitiveString.valueOf("sOmEtHiNg");
CaseInsensitiveString cis6 = CaseInsensitiveString.valueOf("SoMeThInG");
assert cis5 == cis6;
assert cis5.equals(cis6);
Here's the code.
C:\oreyes\samples\java\insensitive>type CaseInsensitiveString.java
import java.util.Map;
import java.util.HashMap;
public final class CaseInsensitiveString {
private static final Map<String,CaseInsensitiveString> innerPool
= new HashMap<String,CaseInsensitiveString>();
private final String s;
// Effective Java Item 1: Consider providing static factory methods instead of constructors
public static CaseInsensitiveString valueOf( String s ) {
if ( s == null ) {
return null;
}
String value = s.toLowerCase();
if ( !CaseInsensitiveString.innerPool.containsKey( value ) ) {
CaseInsensitiveString.innerPool.put( value , new CaseInsensitiveString( value ) );
}
return CaseInsensitiveString.innerPool.get( value );
}
// Class constructor: This creates a new instance each time it is invoked.
public CaseInsensitiveString(String s){
if (s == null) {
throw new NullPointerException();
}
this.s = s.toLowerCase();
}
public boolean equals( Object other ) {
if ( other instanceof CaseInsensitiveString ) {
CaseInsensitiveString otherInstance = ( CaseInsensitiveString ) other;
return this.s.equals( otherInstance.s );
}
return false;
}
public int hashCode(){
return this.s.hashCode();
}
// Test the class using the "assert" keyword
public static void main( String [] args ) {
// Creating two different objects as in new String("Polish") == new String("Polish") is false
CaseInsensitiveString cis1 = new CaseInsensitiveString("Polish");
CaseInsensitiveString cis2 = new CaseInsensitiveString("Polish");
// references cis1 and cis2 points to differents objects.
// so the following is true
assert cis1 != cis2; // Yes they're different
assert cis1.equals(cis2); // Yes they're equals thanks to the equals method
// Now let's try the valueOf idiom
CaseInsensitiveString cis3 = CaseInsensitiveString.valueOf("Polish");
CaseInsensitiveString cis4 = CaseInsensitiveString.valueOf("Polish");
// References cis3 and cis4 points to same object.
// so the following is true
assert cis3 == cis4; // Yes they point to the same object
assert cis3.equals(cis4); // and still equals.
// Lets test the insensitiveness
CaseInsensitiveString cis5 = CaseInsensitiveString.valueOf("sOmEtHiNg");
CaseInsensitiveString cis6 = CaseInsensitiveString.valueOf("SoMeThInG");
assert cis5 == cis6;
assert cis5.equals(cis6);
// Futhermore
CaseInsensitiveString cis7 = CaseInsensitiveString.valueOf("SomethinG");
CaseInsensitiveString cis8 = CaseInsensitiveString.valueOf("someThing");
assert cis8 == cis5 && cis7 == cis6;
assert cis7.equals(cis5) && cis6.equals(cis8);
}
}
C:\oreyes\samples\java\insensitive>javac CaseInsensitiveString.java
C:\oreyes\samples\java\insensitive>java -ea CaseInsensitiveString
C:\oreyes\samples\java\insensitive>
That is, create an internal pool of CaseInsensitiveString objects, and return the corrensponding instance from there.
This way the "==" operator returns true for two objects references representing the same value.
This is useful when similar objects are used very frequently and creating cost is expensive.
The string class documentation states that the class uses an internal pool
The class is not complete, some interesting issues arises when we try to walk the contents of the object at implementing the CharSequence interface, but this code is good enough to show how that item in the Book could be applied.
It is important to notice that by using the internalPool object, the references are not released and thus not garbage collectible, and that may become an issue if a lot of objects are created.
It works for the String class because it is used intensively and the pool is constituted of "interned" object only.
It works well for the Boolean class too, because there are only two possible values.
And finally that's also the reason why valueOf(int) in class Integer is limited to -128 to 127 int values.
In your first example, you are creating a String, "silly" and then passing it as a parameter to another String's copy constructor, which makes a second String which is identical to the first. Since Java Strings are immutable (something that frequently stings people who are used to C strings), this is a needless waste of resources. You should instead use the second example because it skips several needless steps.
However, the String literal is not a CaseInsensitiveString so there you cannot do what you want in your last example. Furthermore, there is no way to overload a casting operator like you can in C++ so there is literally no way to do what you want. You must instead pass it in as a parameter to your class's constructor. Of course, I'd probably just use String.toLowerCase() and be done with it.
Also, your CaseInsensitiveString should implement the CharSequence interface as well as probably the Serializable and Comparable interfaces. Of course, if you implement Comparable, you should override equals() and hashCode() as well.
CaseInsensitiveString is not a String although it contains a String. A String literal e.g "example" can be only assigned to a String.
CaseInsensitiveString and String are different objects. You can't do:
CaseInsensitiveString cis = "Polish";
because "Polish" is a String, not a CaseInsensitiveString. If String extended CaseInsensitiveString String then you'd be OK, but obviously it doesn't.
And don't worry about the construction here, you won't be making unecessary objects. If you look at the code of the constructor, all it's doing is storing a reference to the string you passed in. Nothing extra is being created.
In the String s = new String("foobar") case it's doing something different. You are first creating the literal string "foobar", then creating a copy of it by constructing a new string out of it. There's no need to create that copy.
Just because you have the word String in your class, does not mean you get all the special features of the built-in String class.
when they say to write
String s = "Silly";
instead of
String s = new String("Silly");
they mean it when creating a String object because both of the above statements create a String object but the new String() version creates two String objects: one in heap and the other in string constant pool. Hence using more memory.
But when you write
CaseInsensitiveString cis = new CaseInsensitiveString("Polish");
you are not creating a String instead you are creating an object of class CaseInsensitiveString. Hence you need to use the new operator.
If I understood it correctly, your question means why we cannot create an object by directly assigning it a value, lets not restrict it to a Wrapper of String class in java.
To answer that I would just say, purely Object Oriented Programming languages have some constructs and it says, that all the literals when written alone can be directly transformed into an object of the given type.
That precisely means, if the interpreter sees 3 it will be converted into an Integer object because integer is the type defined for such literals.
If the interpreter sees any thing in single quotes like 'a' it will directly create an object of type character, you do not need to specify it as the language defines the default object of type character for it.
Similarly if the interpreter sees something in "" it will be considered as an object of its default type i.e. string. This is some native code working in the background.
Thanks to MIT video lecture course 6.00 where I got the hint for this answer.
In Java the syntax "text" creates an instance of class java.lang.String. The assignment:
String foo = "text";
is a simple assignment, with no copy constructor necessary.
MyString bar = "text";
Is illegal whatever you do because the MyString class isn't either java.lang.String or a superclass of java.lang.String.
First, you can't make a class that extends from String, because String is a final class. And java manage Strings differently from other classes so only with String you can do
String s = "Polish";
But whit your class you have to invoke the constructor. So, that code is fine.
I would just add that Java has Copy constructors...
Well, that's an ordinary constructor with an object of same type as argument.
String is one of the special classes in which you can create them without the new Sring part
it's the same as
int x = y;
or
char c;
String str1 = "foo";
String str2 = "foo";
Both str1 and str2 belongs to tha same String object, "foo", b'coz Java manages Strings in StringPool, so if a new variable refers to the same String, it doesn't create another one rather assign the same alerady present in StringPool.
String str1 = new String("foo");
String str2 = new String("foo");
Here both str1 and str2 belongs to different Objects, b'coz new String() forcefully create a new String Object.
It is a basic law that Strings in java are immutable and case sensitive.
Java creates a String object for each string literal you use in your code. Any time "" is used, it is the same as calling new String().
Strings are complex data that just "act" like primitive data. String literals are actually objects even though we pretend they're primitive literals, like 6, 6.0, 'c', etc. So the String "literal" "text" returns a new String object with value char[] value = {'t','e','x','t}. Therefore, calling
new String("text");
is actually akin to calling
new String(new String(new char[]{'t','e','x','t'}));
Hopefully from here, you can see why your textbook considers this redundant.
For reference, here is the implementation of String: http://www.docjar.com/html/api/java/lang/String.java.html
It's a fun read and might inspire some insight. It's also great for beginners to read and try to understand, as the code demonstrates very professional and convention-compliant code.
Another good reference is the Java tutorial on Strings:
http://docs.oracle.com/javase/tutorial/java/data/strings.html
In most versions of the JDK the two versions will be the same:
String s = new String("silly");
String s = "No longer silly";
Because strings are immutable the compiler maintains a list of string constants and if you try to make a new one will first check to see if the string is already defined. If it is then a reference to the existing immutable string is returned.
To clarify - when you say "String s = " you are defining a new variable which takes up space on the stack - then whether you say "No longer silly" or new String("silly") exactly the same thing happens - a new constant string is compiled into your application and the reference points to that.
I dont see the distinction here. However for your own class, which is not immutable, this behaviour is irrelevant and you must call your constructor.
UPDATE: I was wrong!
I tested this and realise that my understanding is wrong - new String("Silly") does indeed create a new string rather than reuse the existing one. I am unclear why this would be (what is the benefit?) but code speaks louder than words!
I am working on a section of code for an assignment I am doing atm, and I am completely stuck with 1 little bit.
I need to convert the contents of an array list into a string, or the form of a string, which will be able to be imput into toString() in order for it to be printed to the screen.
public String toString(){
String full;
full = (this.name + this.address + "\n" + "Student Number = " + this.studentId);
for (int i = 0; i < cs.size(); i++) {
full.append(cs[i]);
return full;
The piece of above code is where i attempt to combine 3 varaibles and the contents of an array list into a single string with formatting.
Unfortunatly it creates an error "The type of the expression must be an array type but it resolved to ArrayList"
Thanks for any help.
Jake
cs is array list, so you have to do get operation, not [] (which is for array access)
It should be like:
full.append(cs.get(i));
Not
full.append(cs[i]);
EDIT: As assylis said, full should be StringBuilder not just String, because String doesn't support append() method.
StringBuilder full = new StringBuilder();
Apache Commons StringUtils has different varieties of join() methods that mean you don't have to write this yourself. You can specify the separator and even the prefix/suffix.
I would recommend you look at Apache Commons, not just for this but for lots of other useful stuff.
You are attempting to access an ArrayList as though it is a primitive array (using the square brackets around the index). Try using the get(int index) method instead.
i.e.,
cs.get(i);
You cannot index an ArrayList like an array, you need the get(index) method. Even better, use the enhanced for loop, since it's not recommended to index over a list, as the implementation may change to LinkedList.
I also suggest using a StringBuilder for efficiency:
public String toString() {
StringBuilder full = new StringBuilder();
full.append(this.name);
full.append(this.address);
full.append("\n");
full.append("Student Number = ");
full.append(this.studentId);
for (String s: cs)
full.append(s);
return full.toString();
}
Just use
"cs.get(i)" in place of "cs[i]".
as cs is an ArrayList not an Array.
and also use
full = full + cs.get(i); and not full.append(cs.get(i));
as String type dont have a append method.
Just a note, since you don't put any spacers between each element of the ArrayList it might be unreadable. Consider using Guava's Joiner class.
So instead of
for (...)
s.append(y);
if would be
a.append(Joiner.on(" ").join(yourList));
The Joiner is also more efficient than the for loop since it uses a StringBuilder internally.
I want to ask you about the print vector array , the following one:
Vector[] routingTable = new Vector[connectivity.length];
I tried this method , but it doesn't work with me and it gives me protocol.Route#c17164
when I printed in the main, here is the code, so can you tell me why it doesn't print the correct value ?
public String printRT(int hop)
{
String s = "";
for (int i = 0; i < conf.routingTable[hop].size(); i++)
{
s= " ROUTING TABLE " + conf.routingTable[hop].get(i);
}
return s;
}
it looks like you need to implement the toString() method in protocol.Route.
class Route {
public String toString() {
return "some string that makes sense";
}
}
Either override the toString() method on the protocol.Route class, or get the desired properties from the Route object and append them to the String s inside your printRT method.
Many helpful suggestions, but I think everyone is overlooking something very simple- in each loop iteration you are overwriting the value of s. I think you mean to say something like the following instead:
s += " ROUTING TABLE " + conf.routingTable[hop].get(i);
Note the "+=" rather than simple assignment. Or use a StringBuilder, or whatever.
When you ask java to print an object for which no toString method is defined, then it will fall back on the default toString implementation in the Object class. From the javadocs:
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())
In your example 'protocol.Route' would be the class name and 'c17164' is whatever the hashcode method returns as a hexString, which, unless hashCode has been overwritten, is probably the address of the object, although this is implementation dependent.
So, there are a few ways to fix your problem.
Write your own implementation of the toString method for the Route class that prints out the data you want. This is probably the most "correct" way to fix your problem. It keeps things nicely encapsulated within the class, meaning only the toString method inside of the class needs to know about the exact member variables that are to be printed.
If the situation is such that you cannot change the Route class, you could subclass your own version of the Route class that you could add a toString method to. However, depending on the design of the class, this may be difficult.
Have the current printRT method look inside each Route object and get the specific information that you want to append to the current string.
Also, note that with the current code, you have written the following in the inner loop:
s= " ROUTING TABLE " + conf.routingTable[hop].get(i);
This means that printRT will only return a string for the very last iteration of the loop. So most of the time in the for loop is spent creating strings, assigning them to a variable and then overwriting them the next time through the loop.
If you want to return a string representation for every iteration, you will need to change the above to something like the following:
s += " ROUTING TABLE " + conf.routingTable[hop].get(i);
Now the new information is being appended to s every time through the loop. However, depending on the number of string concatenations being performed, the StringBuilder class may be a better alternative (see a short summary and tutorial on it here).
Two options.
Either override the toString() method on the protocol.Route class.
public String toString() {
return someMethodorPropertyThatreturnsString;
}
or get the desired properties/methods from the Route object and append them to the String s inside your printRT method.
public String printRT(int hop)
{
String s = "";
for (int i = 0; i < conf.routingTable[hop].size(); i++)
{
s= " ROUTING TABLE " + conf.routingTable[hop].get(i).someMethodorPropertyThatreturnsString;
}
return s;
}
There are a number of issues here.
You should be specifying a type to put in your List with Generics. That way, you will make it more obvious to yourself and others what you are putting into and taking out of your List.
As mentioned by others, your List is a list of protocol.Route objects, not Strings. When you try to add a Route to s, Java doesn't know how to convert it into a String, so it uses the default Object#toString(). Override it in Route to do what you want.
It looks like you'll potentially be doing a lot of appending here. Use a StringBuilder.
It looks to me like printRT(int) should be a method inside of whatever conf is.
You should probably be using a different implementation of List; Vector is not really recommended to use anymore, so take a look at other options like ArrayList.