We Keep Coding

Java ,int x = 5; System.out.println(" x + 5 is " + x + 5); System.out.println("x += 5 is " + x += 5); why second println is mistake?

Java
int x = 5;
System.out.println(" x + 5 is " + x + 5);//correct
System.out.println("x += 5 is " + x += 5);// why wrong?
Even though, these 2 println is including calculation but why second println is error.Thanks

What you are doing causes an error because the + is seen as an operator to seperate parts of the string. Try placing that part between brackets like:
System.out.println("x += 5 is " + (x += 5));
This might fix it as you exclude the + from the string. Hope this helps you a bit, and that I am correct in my statement.

Considering two characters as one in a string

String secondLine = ...E......E..E.E;
String failures = "E";
String passed = ".";
int i = 0;
while ((i = (secondLine.indexOf(failures, i) + 1)) > 0) {
System.out.println(i);
feedbackString += "<strong style='color: red;'><li>Failed: </strong><strong>" + i + "</strong> - " + "out of " + resultString.length() + " tests.<br>";
}
The total number of the tests is the sum of the dots which is = 12. and the E's are the failures which in this case = 4.
The dots are the passes and the E's are the failures. Whenever this is ran and there is a failure, it adds an 'E' in front of the dot which becomes'.E'. I can get the E's on its own but I want a statement that says given a dot comes before the E and then it should print the '.E' and pass it into a variable called failedTest.
The code above has an output of: 4, 11, 14, 16 which is not what I want as its considering each character separately but I want it to consider '.E' as one that is if an E comes after a dot then it should consider it as one. if this should be ran considering the '.E' as failure, the expected output should be 3, 9, 11, 12. Thanks :)

I understand your problem better now: if i am right, you print a ., ran the test and if it do not pass, you will print an E.
The point if the final length of secondLine is the number of test plus the number of error, so the if the first error is the test T-i, the T-i dot would be at the position T-i, its E would be at the position (T-1) + 1. The following test would be displaced one position: the dot of the test T-(i+1) would be at the position (T-i) + 2. Do you see the point?
So, this is my suggestion
String secondLine = "...E......E..E.E";
String failures = "E";
String passed = ".";
int detectedErrors = 0;
int i = 0;
while ((i = (secondLine.indexOf(failures, i) + 1)) > 0) {
System.out.println(i);
feedbackString += "<strong style='color: red;'><li>Failed: </strong><strong>"
+ (i - detectedErrors) // here
+ "</strong> - " + "out of " + resultString.length() + " tests.<br>";
detectedErrors += 1; // and here
}

You need to keep track of the number of extra characters in the secondLine and subtract them from the match position. Also change your failure String to ".E"
String secondLine = "...E......E..E.E";
String failures = ".E";
String passed = ".";
int n = 0;
int i = 0;
while ((i = (secondLine.indexOf(failures, i) + 1)) > 0) {
System.out.println(i-n);
n++;
feedbackString += "<strong style='color: red;'><li>Failed: </strong><strong>" + (i-n) + "</strong> - " + "out of " + resultString.length() + " tests.<br>";

Java mysterious output from additions: some evaluated, some printed

I was doing java assignment and I ran into this code.
int x=3, y=5;
System.out.println(x + y + " = " + y + x);
and the output is "8=53". Why does the first x+y gets evaluated and the last y and x expression gets printed? Left me wondering. Thanx in advance guys.

Remember that in Java, an operator (like +) can be overloaded. That means it will do different things, depending on its operands. For +, there are (at least) two choices: integer addition and string concatenation.
Which overload is chosen depends more so on the left-hand-side operand. Also, string concatenation with a non-string operand can cause automatic conversion to a string.
The whole thing will be evaluated left-to-right like this:
x + y + " = " + y + x
3 + 5 + " = " + 3 + 5 // 3+5 chooses the integer addition overload of +
8 + " = " + 3 + 5 // 8 + " = " chooses string concatenation
"8" + " = " + 3 + 5 // so 8 is converted to "8" first
"8 = " + 3 + 5 // and then concatenated with " = "
"8 = " + "3" + 5 // now 3 is converted to "3"
"8 = 3" + 5 // and concatenated with "8 ="
"8 = 3" + "5" // finally 5 is converted to "5"
"8 = 35" // and concatenated with the rest
FWIW, it's ambiguity like this that leads me to dislike implicit conversions1.
1 - In C#. I love it in Python :-)

output is "8=53". Why does the first x+y gets evaluated and the last
y and x expression gets printed?
Because first x + y are not appended with a string so integer calculation is done. Second ones are appended to a string because of " = " + and hence they are printed as individual values.
To prove this just do:
System.out.println("" + x + y + " = " + x + y);
and the output will be:
35 = 35

The first part is getting interpreted as integer addition, but in the second part (as you have introduced a string) it is getting interpreted and string concatenation.
Assuming you want 8 = 8
try
System.out.println(x + y + " = " + (y + x));
if you want 3 + 5 = 5 + 3
then I would use String.format as
in
System.out.println (String.format ("%d + %d = %d + %d", x, y, y, x));

The reason the output is "8 = 53" is because:
x + y + " = " + y + x
Is calculated from left to right so, if we break it down piece by piece:
x + y = 8
8 + " = " = "8 = "
"8 = " + y = "8 = 5"
"8 = 5" + x = "8 = 53"
That is how the compiler gets your answer :)

The first + is between two numbers so the result is eight. The 2nd have strings on either side of them so numbers get converted to strings and concatenated together. The plus operator binds tightest to the left, and gets evaluated left to right. If you wanted the last addition to be numerical then the expression should be in brackets ( ).

System.out.println( number1 + "score" + (number1 + number2) + number1 ) result explanation (string vs. numeric addition operator)

public class Test {
public static void main(String[] args) {
int number1 = 4;
int number2 = 5;
System.out.println( number1 + "Score:" + (number1 + number2) + number1 );
}
}
The output of the above is:
4Score: 94
Why is this? If there is no "score" in there, I understand the result from it, but this I don't know why. number1 and Score: outputs individually first, so then why is it effecting the result to be 94?

You have:
public static void main(String[] args) {
int number1 = 4;
int number2 = 5;
System.out.println( number1 + "Score:" + (number1 + number2) + number1);
}
This is outputting exactly what you told it to output, essentially:
"4" + "Score:" + "9" + "4"
The + operator when used with a string will convert the non-String operands to strings and concatenate the strings together. When + is used with all numeric operands, it is an arithmetic + and just adds the values together.
By putting (number1 + number2) in parentheses, you cause that to be evaluated first, and since both operands are integers, it behaves as an arithmetic + thats adds those two numbers together (producing 9). That result is then converted to a string and concatenated to everything else. It's essentially a shortcut for:
int number1 = 4;
int number2 = 5;
System.out.println( Integer.toString(number1) +
"Score:" +
Integer.toString(number1 + number2) +
Integer.toString(number1) );
If you remove "Score:", then all of the operands are integers, and so all of the + operators are arithmetic addition, and it just sums all the numbers -- i.e. a shortcut for:
System.out.println( Integer.toString(number1 + (number1 + number2) + number1) );
If you want more technical details of the + operator as related to strings vs. numbers, see Section 15.18 of the JLS (15.18.1 describes behavior for strings, 15.18.2 describes behavior for numeric types).
As an aside, the + operator is always left-associative no matter what types the operands are (described in 15.18.1). So the result of the following may surprise you:
System.out.println(1 + 2 + "string" + 1 + 2);
Spoiler (mouse over):
3string12
See http://ideone.com/P11aMI for some more working examples.

It's to do with operator precedence and the overloading of the + operator. If both sides are numbers, then the + operator performs addition:
number1 + number2
results in 9, which is evaluated first (as it's in brackets).
Then as the rest of it has the same precedence, it is overloaded to string concatenation, in left to right order. If one or all of the arguments to the operator are not numbers, they will all be implicitly converted into strings. We start off with:
4 + "Score:" + (4 + 5) + 4
As the brackets are evaluated first, we then get this:
4 + "Score:" + 9 + 4
Which becomes
"4Score:" + 9 + 4
Then
"4Score:9" + 4
And your final result will be
"4Score:94"
just as you got.

the subterm number1 + number2 is treated as an integer. So the sum could be calculated: 9. Using it as a string parameter results in an automatic cast to String, therefore the concatenation of strings is used: number1 + "Score:" + "9" + number1

System.out.println( number1 + "Score:" + (number1 + number2) + number1))
This prints out the following values (the one in the bracket results in an arithmetic sum :
4 + "Score:" + (4+5) + 4
which is
4Score:94

System.out.println( number1 + "Score:" + (number1 + number2) + number1));
it will be step by step as
1) System.out.println( 4 + "Score:" + 9 + 4));
2) System.out.println( "4Score:" + 9 + 4));
3) System.out.println( "4Score:9" + 4));
Then it shows output as
"4Score:94"

Difference between int and int received by ParseInt in java

int i = 0;
int k = Integer.parseInt("12");
int j = k;
System.out.println(i+1 + " " + j+1);
Strangely the output received is
1 121
I can not figure out this basic difference. Please help me.

Use brackets as follows
System.out.println((i+1) + " " + (j+1));
From the docs
The + operator is syntactically left-associative, no matter whether it
is later determined by type analysis to represent string concatenation
or addition. In some cases care is required to get the desired result.
For example, the expression:
a + b + c is always regarded as meaning: (a + b) + c
Extending this to your scenario
i+1 + " " + j+1
it becomes
(((i + 1) + " ") + j)+1
Since i is an int so (i + 1) = 1 , simple addition
" " is a String hence ((i + 1) + " ") = 1 WITH SPACE (String concatenation)
Similarly when j and last 1 is added, its being added to a String hence String concatenation takes place, which justifies the output that you are getting.
See
JLS 15.18.1 String Concatenation Operator +

that is beacuse of " ".
whenever a String comes, java doesnt do any calculations after that and just append it as string.
So in your case, i+1 is computed to 1, but " " + j+1 has string in it. So, it just appended together to form 121

The reason you see this behavior is that the sequence of + operators is evaluated left-to-right. So it is evaluated as if parenthesized:
System.out.println((((i + 1) + " ") + j) + 1);
The first operator adds two int values and produces an int value. The next + adds an int to a String and produces a String. After that, everything is string concatenation. You can introduce your own parentheses to get the result you want.

int i = 0;
int k = Integer.parseInt("12");
int j = k;
System.out.println(i+1 + " " + (j+1));
basically when you put + " " + after this java just appends values as string.
and when you put (j+1) in brackets then its precedence gets higher and it is executes it first and perform sum operation.

When you use " " The expression after that gets evaluated as string.
Using brackets ( and ) around an expression can solve the problem in hand.
System.out.println(i+1 + " " + (j+1));

+ operator is overloaded for addition and String concatenation what you're doing is String concatenation and not addition.. Use brackets for performing addition.

parseint will basically return int (Look at Java API), and there is only one int type in Java. in this example you used " ", where java will treat it as string. in any operation make sure you dont mix up strings with calculations. Always use parenthesis to separate String from calculations.

It happens because the + operator has left associativity and has an overloaded function with strings, so when you have this
int i = 0;
int k = Integer.parseInt("12");
int j = k;
i+1 + " " + j+1
it first sums
i + 1 which gives 1 then it sums 1 + " ", which uses the overloading function of it to concatenate 1 and " " so it gives a string with the value of "1 ". After that it sums "1 " + j and since one of the operands is a string, it does the same behavior and so on.

Interger.parseInt (String str) is a wrapper class method which is used to convert String obj type to primitive data type (int). this are generally used in collection frame work for converting primitive data type to object and vice versa