I am learning Android and taking the MOOC offered by Maryland University.
In one of the lectures I noticed the following line of code:
String output = (val == answer) ? "42" : "We may never know";
My guess is that this is equivalent to:
if(val == answer){
String output = "42";
}else{
String output = "We may never know";
}
Is my assumption correct?
PS: Is there anywhere besides coursera where I can take certified android classes online?
You are almost correct, it's actually like this:
String output;
if(val == answer){
output = "42";
}else{
output = "We may never know";
}
BR Erik
Your assumption is almost correct. The ternary expression would be more accurately represented as:
String output;
if (val == answer) {
output = "42";
}
else {
output = "We may never know";
}
In your original version, output is only available within the scope of the if/else blocks, since you declare it inside them. Declaring it outside the scope means you'll be able to use the value later. However, as Hariharan pointed out, using == for string comparison is a no-no in Java; you're comparing the raw objects, not the string contents. You'll want to replace your val == answer with val.equals(answer) to get a proper result.
Edit: If val and answer aren't strings, disregard the portion about using .equals(). I (and apparently everyone else who answered this question) made the bad assumption that they were, since everything else was strings. .equals() should be used for strings and any other complex object, == can be safely used for primitives.
Yes, you are correct (almost). It's called ternary operator. Your logic is good but you must take care that the variable is initialized outside of the if-else statement.
String output;
if (val == answer) {
output = "42";
} else {
output = "We may never know";
}
If the variable would be defined inside if and else blocks you would not be able to use it after the } sign because it would fall out of the scope.
Yes, you are totally right. It's so called ternary operator. take a look here and here
And a small note: it's weird to declare the same variable twice inside conditional blocks. Declare it somewhere outside .
Related
I'm a beginner in coding. I was recently working with to create a chatting programme where a user will chat with my computer. Here is a part of the code:
System.out.println("Hello, what's our name? My name is " + answer4);
String a = scanner1.nextLine();
System.out.println("Ok, Hello, " + a + ", how was your day, good or bad?");
String b = scanner2.nextLine();
**if (b.equals("good"))** { //1
System.out.println("Thank goodness");
} else **if (b.equals("it was good"))** { //2
System.out.println("Thank goodness");
} else **if (b.equals("bad"))** { //3
System.out.println("Why was it bad?");
String c = scanner3.nextLine();
System.out.println("Don't worry, everything will be ok, ok?");
String d= scanner10.nextLine();
} else **if (b.equals("it was bad"))**{ //4
System.out.println("Why was it bad?");
String c = scanner3.nextLine();
System.out.println("Don't worry, everything will be ok, ok?");
String d= scanner10.nextLine();
}
if(age<18){System.out.println("How was school?");}
else if (age>=18){System.out.println("How was work?");}
The conditions of the if statements are in Bold (surrounded with **). In case of first and the second condition I want my application to do same thing. Similarly third and fourth condition. I thought it was possible to somehow group them in if statement.
I tried with below code but it doesn't compile:
if (b.equals("good"), b.equals("it was good")) {
System.out.println("Thank goodness");
} else if (b.equals("bad"),(b.equals("it was bad"))) {
System.out.println("Why was it bad?");
String c = scanner3.nextLine();
System.out.println("Don't worry, everything will be ok, ok?");
String d= scanner10.nextLine();
}
Can someone correct it for me?
You can use logical operators to combine your boolean expressions.
&& is a logical and (both conditions need to be true)
|| is a logical or (at least one condition needs to be true)
^ is a xor (exactly one condition needs to be true)
(== compares objects by identity)
For example:
if (firstCondition && (secondCondition || thirdCondition)) {
...
}
There are also bitwise operators:
& is a bitwise and
| is a bitwise or
^ is a xor
They are mainly used when operating with bits and bytes. However there is another difference, let's take again a look at this expression:
firstCondition && (secondCondition || thirdCondition)
If you use the logical operators and firstCondition evaluates to false then Java will not compute the second or third condition as the result of the whole logical expression is already known to be false. However if you use the bitwise operators then Java will not stop and continue computing everything:
firstCondition & (secondCondition | thirdCondition)
Here are some common symbols used in everyday language and their programming analogues:
"," usually refers to "and" in everyday language. Thus, this would translate to the AND operator, &&, in Java.
"/" usually refers to "or" in everyday language. Thus, this would translate to the OR operator, ||, in Java.
"XOR" is simply "x || y but both cannot be true at the same time". This translates to x ^ y in Java.
In your code, you probably meant to use "or" (you just used the incorrect "incorrect solution" :p), so you should use "||" in the second code block for it to become identical to the first code block.
Hope this helped :)
You're looking for the "OR" operator - which is normally represented by a double pipe: ||
if (b.equals("good") || b.equals("it was good")) {
System.out.println("Thank goodness");
} else if (b.equals("bad") || b.equals("it was bad")) {
System.out.println("Why was it bad?");
String c = scanner3.nextLine();
System.out.println("Don't worry, everything will be ok, ok?");
String d= scanner10.nextLine();
}
This is probably more answer than you need at this point. But, as several others already point out, you need the OR operator "||". There are a couple of points that nobody else has mentioned:
1) If (b.equals("good") || b.equals("it was good")) <-- If "b" is null here, you'll get a null pointer exception (NPE). If you are genuinely looking at hard-coded values, like you are here, then you can reverse the comparison. E.g.
if ("good".equals(b) || "it was good".equals(b))
The advantage of doing it this way is that the logic is precisely the same, but you'll never get an NPE, and the logic will work just how you expect.
2) Java uses "short-circuit" testing. Which in lay-terms means that Java stops testing conditions once it's sure of the result, even if all the conditions have not yet been tested. E.g.:
if((b != null) && (b.equals("good") || b.equals("it was good")))
You will not get an NPE in the code above because of short-circuit nature. If "b" is null, Java can be assured that no matter what the results of the next conditions, the answer will always be false. So it doesn't bother performing those tests.
Again, that's probably more information than you're prepared to deal with at this stage, but at some point in the near future the NPE of your test will bite you. :)
You can have two conditions if you use the double bars(||). They mean "Or". That means only ONE of your conditions has to be true for the loop to execute.
Something like this:
if(condition || otherCondition || anotherCondition) {
//code here
If you want all of conditions to be true use &&. This means that ALL conditions must be true in order for the loop to execute. if any one of them is false the loop will not execute.
Something like this:
if(condition && otherCondition && anotherCondition) {
//code here
You can also group conditions, if you want certain pairs of them to be true. something like:
if(condition || (otherCondition && anotherCondition)) {
//code here
There is a simpler way.
if (b.contains("good")) {
...
}
else if (b.contains("bad")) {
...
}
Why does java require a double equals sign (==) when comparing Integers in a if statement?
For example
if(x = 3.141)
System.out.println("x is equal to pi.");
is incorrect, it should be
if(x == 3.141)
System.out.println("x is equal to pi.");
I know that "==" is used to compare integers and "=" is used to set an integer value, but why in a if statement does this remain true?
Is it even allowed to assign a variable a value in an if statement (or initiate a new variable)?
Is there any reason anyone would ever want to assign a variable a new value inside an if statement (if so please provide an example)?
This seems like a question that should already have an answer, but I was unable to find one on here or using google, if this is a duplicate question please tell me and I will remove it immediately.
Wouldn't it be confusing if = sometimes did assignment, and sometimes comparison, depending in which context you used it?
That sounds like a bad idea, and would introduce errors.
Plus, the current syntax is compatible with C and C++, so a lot of people are familiar with it.
Is there any reason anyone would ever want to assine a variable a new value inside of an if statement (if so please provide an example)?
It's quite common in while loops:
int b;
while ((b=in.read()) != -1){
=
is used for assignment.
==
is used for comparison.
Is it even allowed to assign a variable a value in an if statement (or initiate a new variable)?
yes it is allowed.
Note what error message you get for if (x = 3.141); it is a type error (cannot convert from double to boolean).
The assignment's type is the type of its both sides; if the type of the assignment is boolean (if (x = true), or even if (x = a.equals(b))), then it is legal to write.
So since it is legal to assign a value to a boolean in the condition, you'd have to use == for comparison.
Is it even allowed to assine a variable a value in an if statement (or initiate a new variable)?
Yes. A common idiom for doing this is:
String line = null;
while ( (line = in.readLine()) != null ) {
// do work
}
In the loop, line is assigned a value and then compared to null. I can't think of an example with ints; it certainly wouldn't be clear there.
History of programming languages 101:
Fortran uses = for both.
Algol introduced := for assignment and used = for comparison. This was required to resolve a grammar ambiguity.
Pascal followed suit.
PL/1 did not.
I can't speak for B or BCPL but by the time we got C it was = for assignment and == for comparison, again to resolve a grammar ambiguity
C++ followed C
Java followed C++ in many respects including this one.
The grammar ambiguity arises because of allowing assignments in expressions. Contrary to your assertion, if (x = true) is legal in Java if x is of type boolean.
== is the identity comparator, which works for both objects and primitives. It answers the question "are the two things the same thing".
= is the assignment operator. It sets the value of the left side to the right side.
Things can turn buggy when using your example with booleans:
boolean b;
if (b = true) // This compiles, but is a bug, because it sets b, not tests it
While other types won't compile with this syntax, boolean and Boolean do, so that's why the following pattern is advised:
if (b)
you can absolutely assign a variable in an if statement. also, that's just the way it works: = always is assignment, and == is always comparison.
So..
= is assignment, and == is comparison, and it is always like this, no matter where they are used.
And assignment is different with "declaration". An assignment statement has its return value, while a declaration doesn't. So you can't write boolean a = false in the () of if statement, but you can write a = false when a has been declared before.
Not all assignments are legal. For example:
int index;
if (index = str.indexOf("something")) {
...
}
It's not legal, because String.indexOf(String) returns an int, while if requires a boolean.
Also, there is a huge difference between "legal" and "making sense".
int index;
if ((index = str.indexOf("something")) != -1) {
...
}
It is legal, as != operation returns a boolean, and it makes sense, as I do want to check if the str contains a substring "something";
However,
int index;
boolean flag;
if ( flag = ((index = str.indexOf("something")) != -1) ) {
...
}
is also legal, as the statement as last returns a boolean; but it DOESN'T make sense, because the != statement already returns a boolean.
I was asked for my homework to make a program wherein the user inputs a Roman numerals between 1-10 and outputs the decimal equivalent. Since I'll be getting a string in the input and an integer in the output, I parsed it, but it won't work. Any ideas why?
import java.util.Scanner ;
class Romans {
static Scanner s = new Scanner(System.in) ;
static String val = null ;
public static void main (String [] args)
{
System.out.print ("Enter a roman numeral between I to X: ");
String val = s.nextLine();
int e = Integer.parseInt(val);
}
static int getRoman (int e)
{
if (val = "I"){
System.out.print ("1") ;
}else if (val = "II" ){
System.out.print ("2") ;
}else if (val = "III") {
System.out.print ("3") ;
} else if (val = "IV") {
System.out.print ("4") ;
} else if (val = "V"){
System.out.print ("5");
} else if (val = "VI") {
System.out.print ("6");
} else if (val = "VII") {
System.out.print ("7");
} else if (val = "VIII") {
System.out.print ("8");
} else if (val = "IX") {
System.out.print ("9");
} else if (val = "X") {
System.out.print ("10") ;
}
return val ;
}
}
Two points:
= is the assignment operator, not the equality-testing operator (==)
You shouldn't use == to test for string equality anyway, as it will only test for reference equality; use equals to test whether two string references refer to equal (but potentially distinct) string objects.
Additionally, you're trying to return a String variable as an int, and you're not even calling getRoman...
I think we can tell you that the correct way to compare Strings is using equals().
You're doing assignments, to compare primitive types you've to use ==, to compare String the equals method.
Example:
if (val.equals("I"))
But also val is not present in the method getRoman().
You are trying to parse val as an int, but its not, its a character.
For such a small sample of chars, its probably easiest to simply create a lookup table, index it on the char.
Are you getting any errors?
In your code, you never call the getRoman function. Also, you're using the assignment operator = instead of the comparison operator "I".equals(val) for example.
String comparsion should be done with equals(String str) method instead of == comparison.
PS. You have = instead of == anyway.
The following statement is an assignment:
val = "I"
That is definitely not what you want to do here.
A comparison is done with the double equals, but double equals (==) compares references but you do not want to do that here either.
You want to use the equals method.
if (val.equals("I")) ...
Make those change everywhere and see how it works for you.
ACtually your main trouble comes from string comparison. In java, = is meant to assign values to variables, == is meant to compare values of primitive types and equals is the way to compare objects, especially for strings.
An alternative to using equals can be to use the JDK internal pool of strings, in this case, you could use == as a comparator.
In your case of parsing roman language numbers, you could also consider using a hashmap to store and retrieve effectively the parsed values of numbers. If you have thousands of comparisons like this to make, then go for identityhashmap.
And last, if you want to do real parsing for all roman numbers, not only the first ones, then you should considering using an automata, i.e. a state machine to parse numbers in a somewhat recursive way, that would be the more efficient model to apply to your problem.
The last 2 remarks are more oriented towards software algorithms, the first two ones are more oriented towards java syntax. You should start to know the syntax before going higher level optimizations.
Regards,
Stéphane
Aside from what was said above about how your String comparison should use the equals( ... ) function - for example,
if ( val.equals("VII") )
you also need to provide a return value for your function called getRoman. This function was declared as a function that returns an integer value to the caller, but in the implementation that you have provided, there are no return values (only System.out.println( ... )).
Also, you aren't inputting the correct parameter type - from what it looks like, your function is checking a String to see if it is a certain Roman numeral. So the correct function header would look like this:
public static int getRoman(String val)
Also, make sure you are actually calling this function in your main() - from what it looks like right now, you aren't even using the getRoman() function.
Hope this helps!
I have an array in containing numbers that represent cable sizes (1, 1.5, 2.5, etc), stored as strings.
In my program, the array is loaded into a spinner, which is working fine.
However, when the item is selected and stored in a variable, I want to check what string was selected, and set another numerical variable to 2.5 so I can do a calculation later in the program.
I tried the following:
if (conductorSize = "1" ) {conCsa = 1;}
else if (conductorSize = "1.5") {conCsa = 1.5;}
conductorSize being the variable holding the selected string, and conCsa being the variable
set to a numerical variable for calculation.
The compiler says that I cannot convert a string to boolean. What's happening?
If you are doing string comparisons, use .equals()
Example taken from here:
String s = "something", t = "maybe something else";
if (s == t) // Legal, but usually WRONG.
if (s.equals(t)) // RIGHT <<<<<<<<<<<<< Use this.
if (s > t) // ILLEGAL
if (s.compareTo(t) > 0) // CORRECT>
As Ed S. points out you are using the assignment operator. However since you are comparing a String you need to use the equals method.
if ("1".equals(conductorSize)) {conCsa = 1;}
else if ("1.5".equals(conductorSize)) {conCsa = 1.5;}
Alternatively, you could just create a new float from your String:
float conCsa;
try {
conCsa = Float.parseFloat(conductorSize);
}catch(NumberFormatException e){
conCsa = 0.0f; //set to a default value
}
It looks like what you're trying to do might better be expressed in this way:
conCsa = Double.parseDouble(conductorSize);
In general you need to use the .equals() method.
If performance is extremely important and you are comparing against string literals, take a look at String.intern(). It'll allow you to do super-fast == comparisons and avoid a full character-by-character scan as in .equals().
Performance would have to be really, really important though, to justify such a non-standard approach.
When you have cable sizes which are constants, you need to use Enums , which will help you in reducing no of if condition comparisons.
I was trying to write a simple method:
boolean validate(MyObject o)
{
// propertyA && propertyB are not primitive types.
return o.getPropertyA() == null && o.getPropertyB() == null;
}
And got a strange error on the == null part:
Syntax error on token ==. Invalid
assignment operator.
Maybe my Java is rusty after a season in PLSQL. So I tried a simpler example:
Integer i = 4;
i == null;
// compile error: Syntax error on token ==. Invalid assignment operator.
Integer i2 = 4;
if (i == null); //No problem
How can this be?
I'm using jdk160_05.
To clarify: I'm not trying to assign anything, just do an && operation between two boolean values. I don't want to do this:
if (o.propertyA() == null && o.propertyB() == null) { return true; }
else { return false; }
== is not an assignment operator, it's a boolean equality operator, see:
http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.21.2
If you want to set i to null use the simple assignment operator =:
i = null;
http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.26.1
If you want to check that i is null then you need to use the == operator
if (i == null)
I don't think you are confusing assignment and equality comparison. I think your compiler is just giving you a confusing error message. This program:
Integer i = 4;
i ==null;
should give an error something like this:
Program.java:8: not a statement
i ==null;
Your first program should compile correctly. Perhaps there is some invisible unicode character in the text that is confusing the compiler. Try deleting the entire function and typing it in again.
I think I see your problem. I'm sorry the other answers don't address it.
So, Java has this idea that is shared by some other languages that just because something is a valid expression doesn't mean that that thing, by itself, is a valid statement.
For example, this code will complain similarly:
Integer i = 4;
i+3; // this line gives a compilation error
And yet obviously I can use i+3 (go unboxing!) elsewhere to mean "7":
System.out.println(i+3); // this is fine
It gets a bit confusing because unlike some languages that have this expression/statement distinction, java allows you to use any method call - whether it returns a value or not - as a statement. However, most java operators do not - by themselves - form a valid statement.
Likewise, this fails to compile:
Integer i = 4;
i; // this line gives a compilation error
For the full gory details, see http://java.sun.com/docs/books/jls/second_edition/html/statements.doc.html#32588
In PL/SQL, assigning a value to a variable is done with the := operator. Comparing two values is done with =.
In Java, assigning a value to a variable is done with the = operator. Comparing two values is done with ==, or a .equals() method in some cases.
You can do things like this:
x = i==null;
This will test if i is null and if so, the value true will be assigned to x (assuming that x is a boolean).