Equality

[Jason Attin]

Hi guys, on Glenn, Mitchell. OCAJP Oracle Certified Associate Java SE 8 Programmer Practice Exams (Kindle Locations 9834-9835). Enthuware. Kindle Edition.
I am not sure why this returns true:

 "hello".concat("        world").trim().equals("hello world");

considering that trim() here doesn't do anything? Are the extra spaces ignored?
thanks

[Henry Wong]

That line of code should return false.

Henry

[Jason Attin]

I would have thought so, but apparently not according to Glenn, Mitchell. OCAJP Oracle Certified Associate Java SE 8 Programmer Practice Exams (Kindle Locations 9021-9027). Enthuware. Kindle Edition. ):

Which of these expressions will return true? Select 4 options
A.

"hello world". equals("hello world")

B.

"HELLO world".equalsIgnoreCase("hello world") 

C.

"hello".concat("       world"). trim().equals("hello world") 

D.

"hello world".compareTo("Hello world") < 0 

E.

"Hello world".toLowerCase( ).equals("hello world")

Correct answers:A,B,C,E.
And here is a screenshot of the question too:

[Henry Wong]

Jason Attin wrote:I would have thought so, but apparently ...

It takes only a few seconds to write a test program to confirm. In fact, you probably could have done it faster than posting this topic...


... and BTW, with the answer, since the "hello" and "world" are on different lines, how do you know that there is only one space?

Henry

[Campbell Ritchie]

Jason Attin wrote:. . . C.

"hello".concat("       world"). trim().equals("hello world") 

. . .

Now, that is interesting. You will have to go to the Java® Language Specification (=JLS). Now, the JLS can be difficult to read at the best of times, but I am going to make you read the grammar. That will be worse. You want the line saying

ExpressionName . [TypeArguments] Identifier ( [ArgumentList] )

Ignore the type arguments and you have
ExpressionName = "hello"
. = . (surprise, surprise ‍)
Identifier = concat
( = (
ArgumentList = "    world"
) = )

Now, it is evident that you have a kind of expression on the left of that line of grammar; you are also saying that it is a kind of expression. So we can fiddle that into

KindOfExpression ::= KindOfExpression . identifier ( argumentList )

So  that grammar shows that the . before identifier associates to the left. This is very important. If the . operator associated to the right, you would execute it like this:-"hello".concat("      world").trim() Rightmost . takes precedence and trim executed "hello".concat("world") Leftmost . executed "helloworld"But . associates to the left, so it is executed like this:-"hello".concat("      world").trim() Leftmost . takes precedence and trim executed "hello    world".trim()) Righttmost . executed. No leading or trailing space to remove. "hello    world"You can show that for yourself very simply:System.out.println("hello".concat("      world").trim());How on earth can you get line C to return true?

[Piet Souris]

I tried it myself yesterday, because like Henry I thought that it should be false. And so it was. Maybe it was a typo, with some parentheses misplaced? This would return true:

"hello".concat(("        world").trim()).equals("helloworld");

But that seems unlikely.

[Paul Anilprem]

This thread turned up while I was searching for something.
This is neither an error nor a typo. It seems some versions of Kindle are stretching a single space a little too much. It shows up correctly on Kindle Fire HDX 8.9 though (tried with Kindle Previewer). The explanation given with this question makes it quite clear that there is a single space between "hello" and "world".

HTH,
Paul.