We Keep Coding

How do I get around the limitation of having to place super constructor in the first line?

public NoWheelsException(Car[] carArray){
String holder = "";
for (int i=0; i<carArray.length; i++) {
if (i == carArray.length - 1) {
holder = holder + carArray[i].name;
}else{
holder = holder + carArray[i].name + ", ";
}
}
String message = holder + " has/have no wheels.";
super(message);
}
Written above is the ideal scenario that I'd have for my code, with the super constructor at the end. Although, since super has to be the first statement, I cannot figure out how to develop the string out of the array inline. I can't straight up use .toString() as there's certain criteria into what the string should look like. I've managed to figure out everything regarding Exceptions except this itty bitty detail. Any help would be greatly appreciated!
Update
I got suggested to try Strin.join in order to link them together although unfortunately the object reference names differ from the name variable in the array objects...

One way is to create a private static method, since static methods exist irrespective of constructors and instantiation:
public NoWheelsException(Car[] carArray){
super(buildMessageFrom(carArray));
}
private static String buildMessageFrom(Car[] cars) {
StringBuilder message = new StringBuilder();
String separator = "";
for (Car car : cars) {
message.append(separator);
message.append(car.name);
separator = ", ";
}
return message.toString();
}
(When building a String in a loop, StringBuilder is much more efficient than string concatenation. Each iteration of ‘holder = holder + …’ would create a new String object that eventually needs to be garbage collected.)
If you’re comfortable with Streams, you can do it all on one line:
public NoWheelsException(Car[] carArray){
super(Arrays.stream(carArray).map(c -> c.name).collect(Collectors.joining(", ")));
}

CodeHS: How use one class's toString method in another class's toString multiple times

This exercise is asking us to make a "RoadTrip" class that creates an ArrayList of geolocations using a geolocation class. Part of the exercise asks that we make a toString method within the RoadTrip Class that would end up returning a string like:
1. San Francisco (37.7833, -122.41671)
2. Los Angeles (34.052235, -118.2436831)
3. Las Vegas (36.114647, -115.1728131)
making a string for each of the GeoLocation objects within the ArrayList.
But I cannot put the return statement in a for loop. Here's an example of me "cheating" to get it do simulate what I would want it do actually do.
public String toString()
{
int counter = 1;
for (int i = 0; i < locationList.size() ; i++)
{
System.out.println(counter + ". " + locationList.get(i).toString());
counter++;
}
return "";
}
If I were to simply replace the System.out.println() with return and remove the return "";, I would get the errors:
RoadTrip.java:43: error: unreachable statement
counter++;
^
RoadTrip.java:45: error: missing return statement
}
^
2 errors
I saw other solutions that would utilize a StringBuilder, but I am assuming that the creators of the curriculum intend that we complete the exercises with the tools we are provided. Is there another method that I can use that would limit itself to the given "toolset"?
Pardon me if my techincal language is off, I'm still relatively new to coding.

Why the problem happens-
The control encounters the return statement on the first loop iteration and goes back to where the method was called from. Hence the following lines in the loop body are not reachable.
Since the return statement is within a loop and is subject to conditional execution, the compiler tells you there is a missing return statement. See code below:
public class Program
{
public static void main(String[] args) {
System.out.println(method());
}
static int method()
{
int i= (int)Math.random();
if(i>0)
return 1;
}
}
Since this is your assignment I won't be providing working code.
The easiest solution would be to define a String variable, store an empty String ("") in it, concat whatever you need in the loop and return it.

If you cannot use StringBuilder, why not concatenate Strings like this;
public String toString()
{
int counter = 1;
String str = "";
for (int i = 0; i < locationList.size() ; i++)
{
str = str + counter + ". " + locationList.get(i).toString();
str = str + "\n";
counter++;
}
return str;
}
P.S - I didn't run the code.

how many time the characters of string are to be found in another string

my professor gave me an exercise to find how many time the characters of string called "filter" are to be found in a second string called "query".
before I begin I am java noob and English isnt my native language.
example:
String filter="kjasd";
String query="kjg4t";
Output:2
getting how many times a char has been found in another string isnt my problem but the problem that the professor gave us some rules to stick with:
class filter. The class must be the following public
Provide interfaces:
public Filter (String letters) (→ Constructor of class)
The string representing the filter should be stored in the letters string
public boolean contains (char character)
Returns true if the passed character is contained in the query string, otherwise false
-public String toString ()
Returns an appropriate string representation of the class (just to be clear I have no clue about what does he means with this one!)
To actually determine the occurrences of the filter in the query, another class QueryResolver is to be created.
The class should be able to be used as follows:
QueryResolver resolver = new QueryResolver();
int count = resolver.where(query).matches(filter).count();
the filter and the query are given by the user.
(i couldnt understand this one! )The methods "where" and "matches" configure the "QueryResolver" to include a subsequent call of "count" the calculation based on the previously passed variables
"query" and "filter" performs.
The count method should use the filter's previously-created method.
The modifier static is not allowed to use!
I dunno if he means that we cant use static {} or we cant use public (static) boolean contains (char character){}
we are not allowed to use void
so the problems that encountered me
- I can not pass a char to the method contains as long as it is not static.
error "Non-static variable can not be referenced from a static context"
i did not understand what i should do with the method toStirng!
what I've done so far:
Approach Nr 1:
so I just wrote everything in the main method to check whether the principle of my code works or not and then I wanted to create that whole with constructor and other methods but unfortunately I did not succeed.
Approach Nr 2:
then I tried to write the code in small mthoden as in the exercise but I did not succeed !.
in both aprroaches i violated the exercise rules but i cant seem to be able to do it alone thats why i posted the question here.
FIRST APPROACH:
public class filter{
public filter(String letters) {
//constructor of the class
String filter;
int count;
}
public boolean contains (char character){
/*Subprogram without static!
*the problem that I can't pass any char to this method if it wasn't static
*and I will get the following error"Non-static variable cannot be referenced from a static context"
*I understand why I'm getting the error but I don't know how to get around it X( */
return true ;
}
public String toString (){
/*he told us to include it in the program but honestly, I don't know what shall I write in it -_-
*I make it to null because you have to return something and I don't know what to do yet
*so, for now, I let it null. */
return null;
}
public static void main(String[] args) {
Scanner in =new Scanner (System.in);
System.out.println("please enter the query string! ");
String query= in.next();
System.out.println("please enter the filter stirng!");
String filter= in.next();
System.out.println("the query string is : [" + query+ "]");
System.out.println("the filter string is : [" + filter+ "]");
int count=0;
// I initialized it temporarily because I wanted to print it!
//later I need to use it with the boolean contains as a public method
boolean contains=false;
//to convert each the query and the filter strings to chars
char [] tempArray=query.toCharArray();
char [] tempArray1=filter.toCharArray();
//to iterate for each char in the query string!
for (int i = 0; i < tempArray.length; i++) {
char cc = tempArray[i];
//to iterate for each char in the filter string!
for (int j = 0; j < tempArray1.length; j++) {
// if the value in the filter string matches the value in the temp array then increment the counter by one!
if(tempArray1[j] == cc){
count++;
contains=true;
}
}
}
System.out.println("the characters of the String ["+filter+"] has been found in the forworded string ["+query+"] exactly "+count+" times!" );
System.out.println("the boolean value : "+ contains);
in.close();
}
}
SECOND APPROACH
- But here too I violated the rules of the task quite brutally :(
- First, I used void and did not use the tostring method.
- Second, I did not use a constructor.
- I did not add comments because that's just the same principal as my first attempt.
public class filter2 {
public static void main(String[] args) {
Scanner in = new Scanner (System.in);
System.out.println("enter the filter string:");
String filterStr=in.next();
System.out.println("enter the query string:");
String querystr =in.next();
Filter(filterStr, querystr);
in.close();
}
public static void Filter(String filterstr , String querystr){
char [] tempArray1 = filterstr.toCharArray();
contains(tempArray1, querystr);
}
public static void contains(char[]tempArray1, String querystr){
boolean isThere= false ;
int counter=0;
char [] tempArray = querystr.toCharArray();
for (int i = 0; i < tempArray.length; i++) {
char cc = tempArray[i];
for (int j = 0; j < tempArray1.length; j++) {
if(tempArray1[j] == cc){
counter++;
isThere=true;
}
}
}
System.out.println("the letters of the filter string has been found in the query string exactly "+counter+" times!\nthus the boolean value is "+isThere);
}
/*
* sadly enough i still have no clue what is meant with this one nor whatshall i do
* public String toString (){
* return null;
* }
*
*/
}
Few hints and advice would be very useful to me but please demonstrate your suggestions in code because sometimes it can be difficult for me to understand what you mean by the given advice. ;)
Thank you in advance.
(sorry for the gramatical and the type mistakes; english is not my native language)

As already mentioned, it is important to learn to solve those problems yourself. The homework is not for punishment, but to teach you how to learn new stuff on your own, which is an important trait of a computer scientist.
Nonetheless, because it seems like you really made some effort to solve it yourself already, here is my solution, followed by some explanation.
General concepts
The first thing that I feel like you didn't understand is the concept of classes and objects. A class is like a 'blueprint' of an object, and the object is once you instanciated it.
Compared with something like a car, the class would be the description how to build a car, and the object would be a car.
You describe what a class is with public class Car { ... }, and instanciate an object of it with Car myCar = new Car();.
A class can have methods(=functions) and member variables(=data).
I just repeat those concepts because the code that you wrote looks like you didn't fully understand that concept yet. Please ask some other student who understood it to help you with that.
The Filter class
public class Filter{
String letters;
public Filter(String letters) {
this.letters = letters;
}
public boolean contains (char character){
for(int i = 0; i < letters.length(); i++) {
if(letters.charAt(i) == character)
return true;
}
return false;
}
public String toString (){
return "Filter(" + letters + ")";
}
}
Ok, let's brake that down.
public class Filter{
...
}
I guess you already got that part. This is where you describe your class structure.
String letters;
This is a class member variable. It is unique for every object that you create of that class. Again, for details, ask other students that understood it.
public Filter(String letters) {
this.letters = letters;
}
This is the constructor. When you create your object, this is the function that gets called.
In this case, all it does is to take an argument letters and stores it in the class-variable letters. Because they have the same name, you need to explicitely tell java that the left one is the class variable. You do this by adding this..
public boolean contains (char character){
for(int i = 0; i < letters.length(); i++) {
if(letters.charAt(i) == character)
return true;
}
return false;
}
This takes a character and looks whether it is contained in this.letters or not.
Because there is no name collision here, you can ommit the this..
If I understood right, the missing static here was one of your problems. If you have static, the function is class-bound and not object-bound, meaning you can call it without having an object. Again, it is important that you understand the difference, and if you don't, ask someone. (To be precise, ask the difference between class, object, static and non-static) It would take too long to explain that in detail here.
But in a nutshell, if the function is not static, it needs to be called on an object to work. Look further down in the other class for details how that looks like.
public String toString (){
return "Filter(" + letters + ")";
}
This function is also non-static. It is used whenever the object needs to be converted to a String, like in a System.out.println() call. Again, it is important here that you understand the difference between class and object.
The QueryResolver class
public class QueryResolver {
Filter filter;
String query;
public QueryResolver where(String queryStr) {
this.query = queryStr;
return this;
}
public QueryResolver matches(String filterStr) {
this.filter = new Filter(filterStr);
return this;
}
public int count() {
int result = 0;
for(int i = 0; i < query.length(); i++) {
if(filter.contains(query.charAt(i))){
result++;
}
}
return result;
}
}
Again, let's break that down.
public class QueryResolver {
...
}
Our class body.
Note that we don't have a constructor here. It is advisable to have one, but in this case it would be an empty function with no arguments that does nothing, so we can just leave it and the compiler will auto-generate it.
public QueryResolver where(String queryStr) {
this.query = queryStr;
return this;
}
This is an interesting function. It returns a this pointer. Therefore you can use the result of the function to do another call, allowing you to 'chain' multiple function calls together, like resolver.where(query).matches(filter).count().
To understand how that works requires you to understand both the class-object difference and what exactly the this pointer does.
The short version is that the this pointer is the pointer to the object that our function currently lives in.
public QueryResolver matches(String filterStr) {
this.filter = new Filter(filterStr);
return this;
}
This is almost the same as the where function.
The interesting part is the new Filter(...). This creates the previously discussed Filter-object from the class description and puts it in the QueryResolver object's this.filter variable.
public int count() {
int result = 0;
for(int i = 0; i < query.length(); i++) {
if(filter.contains(query.charAt(i))){
result++;
}
}
return result;
}
Iterates through the object's query variable and checks for every letter if it is contained in filter. It keeps count of how many times this happens and returns the count.
This function requires that filter and query are set. Therefore it is important that before someone calls count(), they previously call where(..) and matches(..).
In our case, all of that happens in one line, resolver.where(query).matches(filter).count().
The main function
I wrote two different main functions. You want to test your code as much as possible during development, therefore the first one I wrote was a fixed one, where you don't have to enter something manually, just click run and it works:
public static void main(String[] args) {
String filter="kjasd";
String query="kjg4t";
QueryResolver resolver = new QueryResolver();
int count = resolver.where(query).matches(filter).count();
System.out.println(count);
}
Once you understand the class-object difference, this should be straight forward.
But to repeat:
QueryResolver resolver = new QueryResolver();
This creates your QueryResolver object and stores it in the variable resolver.
int count = resolver.where(query).matches(filter).count();
Then, this line uses the resolver object to first call where, matches, and finally count. Again, this chaining only works because we return this in the where and matches functions.
Now finally the interactive version that you created:
public static void main(String[] args) {
Scanner in =new Scanner(System.in);
System.out.println("please enter the query string! ");
String query= in.next();
System.out.println("please enter the filter stirng!");
String filter= in.next();
System.out.println("the query string is : [" + query+ "]");
System.out.println("the filter string is : [" + filter+ "]");
QueryResolver resolver = new QueryResolver();
int count = resolver.where(query).matches(filter).count();
System.out.println("the characters of the String ["+filter+"] has been found in the forworded string ["+query+"] exactly "+count+" times!" );
in.close();
}

Write a recursive method that returns the number of occurences of 'A' in the string it is passed

So I've been fiddling with this problem for the past hour. I keep getting unexpected type error. It appears to be from a confliction between charAt and s.length. Any ideas on what could fix that?
class lab7
{
public static void main(String[] args)
{
String s = ("BCA");
}
public static String recursion(String s)
{
if (s.length()>=0)
{
if(s.charAt(s.length()) = A)
{
count++;
}
s.substring(0, s.length()-1);
}
return count;
}
}

There are several issues with this code, including some significant logic errors. However, the specific error you're getting is probably here:
if(s.charAt(s.length()) = A)
First, note that you're using = instead of ==, which does an assignment rather than a comparison. Also note that A should be in single quotes to be a character literal. Right now, Java thinks A is the name of a variable, which isn't defined. Finally, note that strings are zero-indexed, so looking up the character at position s.length() will give you a bounds error.
I hope this helps you get started! As a hint, although your function is named "recursion," does it actually use recursion?

Following code uses String class. For performance critical applications you might want to use StringBuffer / StringBuilder class accordingly.
class StringCounter
{
public static void main (String[] args)
{
int count = returnCount("ABCDABCDABCD", 0);
System.out.println(count);
}
public static int returnCount(String s, int count)
{
// You may want to do some validations here.
if(s.length()==0)
{
return count;
}
if(s.charAt(0)=='A')
{
return returnCount(s.substring(1), count+1);
}
else
{
return returnCount(s.substring(1), count);
}
}
}
The code simply slices the String parameter one character at a time and checks for the required character. Further on every invoke it will update the count and String parameter.
Any ideas on what could fix that?
Your function is not recursive. Recursive functions call themselves with manipulated/updated parameters.
As a thumb rule in recursive functions, always think in terms of manipulating function parameters.
Always have a base case that will terminate recursive calls.

Consider this snippet:
static int countA(String str) {
if (str == null || str.length() == 0) { /* nothing or "" contains 0 A's */
return 0;
}
return (str.charAt(0) == 'A' ? 1 : 0 ) /* 0 or 1 A's in first character */
+ countA(str.substring(1)); /* plus no. of A's in the rest */
}
And you call the function like this:
int a = countA("ABAABA"); /* a is 4 */
I realize now that this question was school related, but at least this snippet works as an exercise in understanding recursion.

Insert string in beginning of another string

How to insert a string enclosed with double quotes in the beginning of the StringBuilder and String?
Eg:
StringBuilder _sb = new StringBuilder("Sam");
I need to insert the string "Hello" to the beginning of "Sam" and O/p is "Hello Sam".
String _s = "Jam";
I need to insert the string "Hello" to the beginning of "Jam" and O/p is "Hello Jam".
How to achieve this?

The first case is done using the insert() method:
_sb.insert(0, "Hello ");
The latter case can be done using the overloaded + operator on Strings. This uses a StringBuilder behind the scenes:
String s2 = "Hello " + _s;

Other answers explain how to insert a string at the beginning of another String or StringBuilder (or StringBuffer).
However, strictly speaking, you cannot insert a string into the beginning of another one. Strings in Java are immutable1.
When you write:
String s = "Jam";
s = "Hello " + s;
you are actually causing a new String object to be created that is the concatenation of "Hello " and "Jam". You are not actually inserting characters into an existing String object at all.
1 - It is technically possible to use reflection to break abstraction on String objects and mutate them ... even though they are immutable by design. But it is a really bad idea to do this. Unless you know that a String object was created explicitly via new String(...) it could be shared, or it could share internal state with other String objects. Finally, the JVM spec clearly states that the behavior of code that uses reflection to change a final is undefined. Mutation of String objects is dangerous.

Sure, use StringBuilder.insert():
_sb.insert(0, _s);

You can add a string at the front of an already existing one. for example, if I have a name string name, I can add another string name2 by using:
name = name2 + name;
Don't know if this is helpful or not, but it works. No need to use a string builder.

private static void appendZeroAtStart() {
String strObj = "11";
int maxLegth = 5;
StringBuilder sb = new StringBuilder(strObj);
if (sb.length() <= maxLegth) {
while (sb.length() < maxLegth) {
sb.insert(0, '0');
}
} else {
System.out.println("error");
}
System.out.println("result: " + sb);
}

import java.lang.StringBuilder;
public class Program {
public static void main(String[] args) {
// Create a new StringBuilder.
StringBuilder builder = new StringBuilder();
// Loop and append values.
for (int i = 0; i < 5; i++) {
builder.append("abc ");
}
// Convert to string.
String result = builder.toString();
// Print result.
System.out.println(result);
}
}

It is better if you find quotation marks by using the indexof() method and then add a string behind that index.
string s="hai";
int s=s.indexof(""");