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Why printing "" + variables output each variable individually without spacing?

One of my assignments has this code:
System.out.println("" + x + y + count);
that outputs the value of x, y and count individually without any spaces. I would like to know more about it online. However, I can't seem to find the right keywords to search it up online. Can someone please explain to me the logic behind this or perhaps point to me a name or keyword for such a situation?
I have always known the " " as a tool to print out a string so I'm confused by this.
Thanks in advance.

If we apply standard Java precedence rules, the statement:
System.out.println("" + x + y + count);
is equivalent to
System.out.println((("" + x) + y) + count);
Then we look at the meaning of +
If the static types of both a and b are numeric types (either primitive numeric or their boxed types) then a + b is numeric addition.
Otherwise, a + b is string concatenation. The two arguments are converted to strings and the strings are concatenated.
Based on this we can say that all of the + operators in the example will be treated as string concatenations.
If you want spaces between x, y and count you need to add some string literals; e.g.
System.out.println("" + x + " " + y + " " + count);
or, more simply:
System.out.println(x + " " + y + " " + count);
If you wanted x, y and count to be added (assuming that they are numeric), then you could write this:
System.out.println("" + (x + y + count));
or, more simply:
System.out.println(x + y + count);
The latter is using a different overload of println.
I have always known the "" as a tool to print out a string so I'm confused by this.
Ummm ... it is actually an empty string literal. The usage "" + x is simply an idiom for converting x to a String. The empty string literal has other uses too. The main one is to represent a String with zero characters.

Or you can use String format like this:
System.out.printf("%d%d%d", x, y, count);

The "" is an empty String. In Java when you concatenate a String with other primitives that can be cast to String the result is a String. That means that the code
System.out.println("" + something);
is a different way to write
System.out.println(String.valueOf(something));
However in your scenario the "" + x + y + count means that the elements are converted to String and then concatenated - this means that if x==1, y==2, count==3 the result would be 123. If you wanted to just cast the result to String you would have to indicate that the computation should happen before casting to String for example by using brackets
System.out.println("" + (x+y+count));
The output of this would be 6.

Java: Issue when replacing Strings on loop

I'm building a small app which auto translates boolean queries in Java.
This is the code to find if the query string contains a certain word and if so, it replaces it with the translated value.
int howmanytimes = originalValues.size();
for (int y = 0; y < howmanytimes; y++) {
String originalWord = originalValues.get(y);
System.out.println("original Word = " + originalWord);
if (toReplace.contains(" " + originalWord.toLowerCase() + " ")
|| toCheck.contains('"' + originalWord.toLowerCase() + '"')) {
toReplace = toReplace.replace(originalWord, translatedValues.get(y).toLowerCase());
System.out.println("replaced " + originalWord + " with " + translatedValues.get(y).toLowerCase());
}
System.out.println("to Replace inside loop " + toReplace);
}
The problem is when a query has, for example, '(mykeyword OR "blue mykeyword")' and the translated values are different, for example, mykeyword translates to elpalavra and "blue mykeyword" translates to "elpalavra azul". What happens in this case is that the result string will be '(elpalavra OR "blue elpalavra")' when it should be '(elpalavra OR "elpalavra azul")' . I understand that in the first loop it replaces all keywords and in the second it no longer contains the original value it should for translation.
How can I fix this?
Thank you

you can sort originalValues by size desc. And after that loop through them.
This way you first replace "blue mykeyword" and only after you replace "mykeyword"

The "toCheck" variable is not explained what is for, and in any case the way it is used looks weird (to me at least).
Keeping that aside, one way to answer your request could be this (based only on the requirements you specified):
sort your originalValues, so that the ones with more words are first. The ones that have same number of words, should be ordered from more length to less.

Why does the expression x+x not print the same result in the two places it appears? [duplicate]

This question already has answers here:
Java: sum of two integers being printed as concatenation of the two
(10 answers)
Closed 7 years ago.
Why does the expression x+x not print the same result in the two places it appears?
String s = args[0];
System.out.println("Hello "+s);
int x = 40;
System.out.println(x);
System.out.println(x+x);
System.out.println(s+" "+x+x);
The result of this code is when I execute in cmd java EG1 kaan
Hello kaan
40
80
kaan4040
why is the last result of the print displaying kaan4040 and not kaan80?

Because of automatic conversion to String.
On this line you "start printing" an integer, so adding another integer to it will again produce integer that is then converted to String and printed out:
System.out.println(x + x); // integer + integer
However on this line you "start printing" a String, so all other values you add to it are at first converted to String and then concatenated together:
System.out.println(s + " " + x + x); // String + String + integer + integer
If you want the two integers to be added together before the concatenation is done, you need to put brackets around it to give it a higher priority:
System.out.println(s + " " + (x + x)); // String + String + integer

In your last print statement, you are doing a string concatenation instead of an arithmetic addition.
Change System.out.println(s+" "+x+x) to System.out.println(s+" "+(x+x)).

Make changes System.out.println(s+" "+x+x); to System.out.println(s+" "+(x+x)); Because it need to add the value and then string concatenation

Because java does some work with your code. When you do System.out.println(x+x);, it sees x+x as an expression with two ints and evaluates it (which is 80). When you do ""+x+x, it sees 3 String, and thus evaluates this expression as a String concatenation.
(btw, by it, I mean javac, and "sees", I mean, well "reads")

Or change print code to System.out.println(x +x+" " +s );

You are performing concatenation instead of addition

Whenever you append anything to string then it will result to string only. You have appended x+x to " " which will append 40 after name. You can use System.out.println(s+" "+(x+x)).

On the last print statement:
System.out.println(s+" "+x+x);
s is a string and is concatenated with " ", from left to right the expression formed by concatenation with s and " ", is then concatenated with x and then ( s + " " + x ) is concatenated with x, yielding kaan4040.
If the + operator is used with:
2 Strings, concatenation occurs
1 String and 1 int, concatenation occurs
2 ints, arithmetic addition
Consider the following scenario:
System.out.println(x + x + " " + "hello");
In this example 80Kaan is printed as arithmetic addition occurs between x and x, then the resulting value (80) is concatenated with the space and hello.
Read from left to right.

int x = 40;
System.out.println(x);
System.out.println(x + x);
System.out.println("" + x + x);
40
80
4040
40 is int 40
80 is int 40 + int 40 (Maths)
4040 is String 40 concat String 40 (because add "" String)

String s = args[0];
System.out.println("Hello "+s);
int x = 40;
System.out.println(x); //1st statement
System.out.println(x+x); //2nd statement
System.out.println(s+" "+x+x); //3rd statement
The first statement simply converts x into String
The second satatement added the numbers because there aren't strings, the compiler thinks of plus sign as addition of two numbers.
the third one sees that there is a string so the compiler thinks like:
print the value of s, add space(" "), add the value of x (convert x into string), add the value of x (convert x into string ).
Hence, Kaan4040.
If you want to print 80, you can do it in two ways:
int sum = x+x;
System.out.println(s+" "+sum); //more readable code
or
System.out.println(s+" "+ (x+x) ); //may confuse you
the compiler will think of x+x as integers since it doesn't find any string inside the parenthesis. I prefer the first one though. :)

why is the last result of the print displaying kaan4040 and not kaan80?
This is because this is how String behaves when used with the + symbol. and it can mean differently when used in a println method.
It means String concatenation when you use it with a String
The 5 even though being an integer will be implicitly converted to String.
E.g:
System.out.println("Hello" + 5); //Output: Hello5
It become mathematical operation:plus when used within a pair of brackets because the brackets will be operated first (add first), then convert to String.
The 1st + is concatenation, and 2nd + is add (Refer to codes below).
E.g:
System.out.println("Hello" + (5+7)); //Output: Hello12

If any one of the '+' operator operand is string, then java internally create 'StringBuilder' and append those operands to that builder. for example:
String s = "a" + 3 + "c";
it's like
String s = new StringBuilder("a").append(3).append("c").toString(); //java internally do this

length of a String with surrogate characters in it - java

I am having trouble counting the length of my String which has some surrogate characters in it ?
my String is,
String val1 = "\u5B66\uD8F0\uDE30";
The problem is, \uD8F0\uDE30 is one character not two, so the length of the String should be 2.
but when I am calculating the length of my String as val1.length() it gives 3 as output, which is totally wrong. how can I fix the problem and get the actual length of the String?

You can use codePointCount(beginIndex, endIndex) to count the number of code points in your String instead of using length().
val1.codePointCount(0, val1.length())
See the following example,
String val1 = "\u5B66\uD8F0\uDE30";
System.out.println("character count: " + val1.length());
System.out.println("code points: "+ val1.codePointCount(0, val1.length()));
output
character count: 3
code points: 2
FYI, you cannot print individual surrogate characters from a String using charAt() either.
In order to print individual supplementary character from a String use codePointAt and offsetByCodePoints(index, codePointOffset), like this,
for (int i =0; i<val1.codePointCount(0, val1.length()); i++)
System.out.println("character at " + i + ": "+ val1.codePointAt(val1.offsetByCodePoints(0, i)));
}
gives,
character at 0: 23398
character at 1: 311856
for Java 8
You can use val1.codePoints(), which returns an IntStream of all code points in the sequence.
Since you are interested in length of your String, use,
val1.codePoints().count();
to print code points,
val1.codePoints().forEach(a -> System.out.println(a));

Why my code is giving NumberFormatException?

I have code like this :
int [] arrayOfImages = new int[namesOfSubjectsColorCode.size()];
int y = 0;
for (int x = 0 ; x<namesOfSubjectsColorCode.size();x++) {
nameOfColorCode = namesOfSubjectsColorCode.get(x);
String str = "com" + "." + "nyurals" + "." + "R" + "." + "drawable" + "." + nameOfColorCode;
arrayOfImages[y] = Integer.parseInt(str);
// Integer.parseInt(str);
y++;
}
Here, I have created integer array. Then, I have created string and by using Integer.parseInt() I want to convert it to int so that, my array of integer should generate dynamically. It is giving NumberFormatException.
Please suggest to me a solution for this.

There is no way for this String to be reasonably turned into an int.
String str = "com" + "." + "nyurals" + "." + "R" + "." + "drawable" + "." + nameOfColorCode;
Something like this would be expected:
String str = "1";

Its obvious that it gives you NumberFormatException. Look at your code :
Your str variable contains String which cant parse into Integer value.

Argument for Integer.parseInt() is invalid, you can't pass it string like "com.nyurals.." etc
From the docs:
public static int parseInt(String s)
throws NumberFormatException
Parses the string argument as a signed decimal integer. The characters in the string must all be decimal digits, except that the first character may be an ASCII minus sign '-' ('\u002D') to indicate a negative value or an ASCII plus sign '+' ('\u002B') to indicate a positive value. The resulting integer value is returned, exactly as if the argument and the radix 10 were given as arguments to the parseInt(java.lang.String, int) method.
And that's exactly what you're getting: NumberFormatException.
EDIT:
You probably want to do something like this:
nameOfColorCode = namesOfSubjectsColorCode.get(x);
String str = "" + nameOfColorCode;
int resourceId = this.getResources().getIdentifier(str, "drawable", this.getPackageName());
arrayOfImages[y] = resourceId;
y++;

If you want to get the int (id) of a resource than you should use
Resources res = activity.getResources();
res.getIdentifier("name","resourceType",activty.getPackageName());
change the name with actual name and resourceType with actual resource type (drawable,color,etc);
As others wrote you don't have an int in your string, it's just an int and not a reference to resoruce

You can access the constants you're trying to reference via http://docs.oracle.com/javase/tutorial/reflect/ but you really don't want to do that.
You'd be better served putting the int's in some form of map of which you can then select some kind of subset dependent on changing information

Integer.parseInt() expects to be given an integer of some sort, not a string of the format com.nyurals.R.drawable.<<nameOfColorCode>>.
Generally, in Android, you would access the variables directly with something like:
R.drawable.SomeName
rather than trying to decode (or, in your case, directly using) the string to get a value.
For example, the following code gets the surface view based on an ID:
SurfaceView sv = (SurfaceView)findViewById(R.id.PfrRightAntiView);
If you do need to dynamically extract a resource based on a string known only at runtime, look into Resources.getIdentifier(), an example of which can be found here.
As per this article, that method may not be the fastest. It may be better to use the reflection method as shown in that link.

May your str returns null so Integer.parseInt(null) return number format exception