Question about double brackets for boolean condition (possibly Java)? - java

I have a vague memory of my professor from way back ~10 years ago saying that it would be a good practice to write conditions which check a boolean value, with double brackets, unless if you specifically check if it is true or is false.
So instead of
if(isColdOutside) {}
You should do
if((isColdOutside)) {}
And the reason is that if you do single-bracket, the compiler is going to issue a warning to tell you that maybe you forgot to check the value, since there is no == in the condition. So by using double-brackets, you tell the compiler "This is what i meant to do, just check if this is truthy".
He mostly taught Java stuff, that's why I thought it might be a Java thing.
But now, I wanted to look for it, and I can't find anything about this. I tried searching for variations of stuff like "double bracket condition" or "boolean condition warning", and I got no significant results.
Does this syntax/convention have a name? Is it even real, or did I just dream it?

After 20 years of working with Java I have never heard of this. I don't think it is a thing. And if it is a thing, then no one I know would think it was something other than a typo.
Other languages like the Bash shell in linux have multi-bracket differences, but Java, not so much.

Related

Does bitwise inclusive OR with assignment operator have advantages over typical assignment in java?

When looking at some code on one or another git, sometimes I can see that devs use bitwise inclusive OR compound assignment operator (|=) where simple assignment would be enough. Unfortunately, I don't have any code with this solution at hand, so I'll try to describe it as best I can.
Let's say, we have the following code in java:
boolean found = false;
for (String s : stringSet) {
if (s == null || s.equals("")) {
found |= true; // <== this line
}
}
if (!found) {
System.out.println("No interesting string found");
}
I ask only about the pointed line. Why people do it this way? I understand that we can have a really great amount of data, billions or trillions to iterate over. But does the pointed line changes the efficiency so dramatically? Would it be noticeably slower for a lot of data, if I change it to simple assignment: found = true;?
I don't exclude the possibility that not a speed is the main argument, but it seemed most meaningful to me.
And yes, I know this code can be converted to method or streams, but it's only a simplification of a code where it would be far more complicated etc. We can assume that before the pointed line (or even before if), there are tons of code that do something meaningful. Please, don't suggest something like "use streams instead", because I know how to java advanced already. I'd like to understand the phenomenon of this somehow enigmatic solution using bitwise inclusive OR.

Reverse words in Java without using any language specific functions

I have been going through the Java interview questions asked by my company and came across one that I can't seem to find the solution.
Here is the question:
Please write a method (function) accepting as single parameter a
string and reversing the order of the words in this string.
The " " is the word separator and any other char is considered as being part of a word. In order to simplify, please consider that there is always one space between the words.
Important - You are NOT allowed to use other strings or arrays or other data structures containing several elements - just plain atomic variables such as integers, chars etc.
Also, it is not allowed to use any other language specific string function other than the function giving you the length of the string.
Expected result:
"hello my beautiful world" -> "world beautiful my hello"
So, I can't use: chars[], str.split(), str.charAt(), str.substring(), StringBuilder, another declaration of String.
Should I use recursion to do it?
Since, String is Immutable and uses encapsulation,
There is no solution to your problem. You can't update the values directly, no setters are available and without the access to the getters (since you can only use .length), you can't read the value.
So, I would suggest to respond that Immutability and encapsulation prevent you from doing so.
In real life as a software engineer, you'll sometimes be asked to do things that are technically impossible or even nonsensical. Sometimes the person asking will be someone important like your boss or a big customer.
If someone actually asks you this interview question, then you're in one of those situations. That makes this question pretty interesting, and you might want to figure out what the best way to answer really is.
If someone asked me, this is how I would answer, and as an interviewer, this is the kind of answer I would award the most points for:
1) Explain how it's technically impossible to meet the requirements, but do it without making me feel stupid. This shows diplomacy.
2) Figure out what I really want. In this case, the interviewer probably wants to see if you know how to reverse the words in a string using low-level operations. This is a perfectly reasonable C language question, for example. Figuring out what the interviewer really wants shows experience and judgement.
3) Provide an answer that gives me what I want. Write this method in Java, but take a StringBuilder instead of a string, and call only length(), charAt(), and setCharAt(). This shows the expertise that the interviewer wants to see.

Code with i+= in the loop is not working

I have the following piece of code called Code1.
http://pastebin.com/tc0Vd8xh
When I run this, the sketch does not work.
However when I replace "i=+50" for "i = i + 50" the code works.
My question is why the "i=+50" bit does not work?
As far as I know "i=+50" is proper Java and Processing is based on Java.
I tried to Google about "i=+50" but Google does not process non-alphanumeric characters.
So I came here and I searched in previous questions before asking here. Anyone, any idea why "i=+50" does not work?
The statement i=+50 is the assignment of positive 50 to i. That is why it compiles, but doesn't add 50 to i on each loop. As #RoelHarbers and #ByoTic mentioned, you actually want i += 50
You're using =+, which is not a java operator (or an operator in any other language I know of)
The proper syntax is:
i+=50
Because it's i+=50 and not i=+50.
i =+ 50 is not going to do what you want, it is going to initialize i with 50. Instead, use i+=50, this going to add the 50 to whatever value that i holds.

Java one line if statement [duplicate]

This question already has answers here:
Is it ok if I omit curly braces in Java? [closed]
(16 answers)
Closed 9 years ago.
I am using if condition without braces in java something like
if(somecondition)
//Only one line Business logic
but some told use braces always even one line statement something like this
if(somecondition){
//Only one line Business logic
}
What is the better way according to java sandard?
there's no real "standard". i prefer always using braces because if you dont you risk someone adding an innocent looking logging statement turning your code from
if(somecondition)
//Only one line Business logic
into
if(somecondition)
log.debug("condition was true");
//Only one line Business logic
and then things stop working :-)
That's a matter of taste. I would use braces or else no braces but write all code in one line to improve readability.
Also you might consider using a ternary operator
booleanExpression ? value1 : value2
In addition to #radai answer, if you are a real evil mind, when you see a if with no braces you can do something that will make you ennemies by adding a semi-colon on the same line of the if but at the 800th column of the line(or something).
like
if(condition) /*a loooot of whitespace*/ ;
//Only one line Business logic that will get executed whatever is the condition
This is why i prefer to use braces and recommend people to use them
No naked if statements. You're just asking for trouble. Always use { }
it is better to use braces when checking for errors or updating the code.
imagine.
if(answer.equals("add"))
addedValue += Scanner.readInt();
but you have a new requirement to add only the absolute value, so you change to.
if(answer.equals("add2))
valueToBeAdded = Scanner.readInt();
if(valueToBeAdded < 0) valueToBeAdded = - valueToBeAdded;
addedValue += valueToBeAdded;
this is not a really correct algorithm, is just an example of what can happens.
Using if statement with braces is better way to java standard, because it increase the readability and reduce unwanted error.
The two statements have exactly the same effect but I have suffered so often from the lack of braces that I also always comment that there should be braces even around 1 line statements. This makes the code easier to maintain and can save a lot of headache. My experience shows that one line if statements often turn into multi-line statements on later iterations so what you save by not writing two { the first time, you will give later on.
According to java standard braces are better because if they are not there compiler has to work around more and also would be performance issue.

How to have an IF OR in android?

if(word.equals(" ")){
}
I have this IF statment, what i would like it to also do is the following..
if(word.equals(" ") OR word.equals(".") ){
}
I know the above code will not work, but im looking how to implement something like that? Anyone know how to do an if or, or do i have to use else if?
Android uses Java, and the boolean OR operator in Java (and in pretty much every single language with a C-inspired syntax) is ||.
if (word.equals(" ") || word.equals("."))
Here is a document describing the Java operators. You may also want to read it from the start since it covers the basics of Java.
There are thousands of tutorials on Google explaining the basics of the Java language that you could also use.
Probably, the reason the question was asked is that you must do this test BEFORE trying to do a .parseDouble on 'word' or else the app will crash

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