can you have two conditions in an if statement - java

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")) {
...
}

Related

Cleaner way to write code snippet

I'm new to java and I was wondering if there was an easier way to write
if(a == 10 || b == 10){
//stuff
}
In my mind I tried something like this:
if(a||b == 10){
//stuff
}
because IMO that makes a lot of intuitive sense, but it's not a thing.
if you're only comparing a few values then you might as well proceed with the current approach as there is nothing in place to make it shorter. However, if you're repeating your self many times, then you can create a helper function to do the work for you.
i.e
static boolean anyMatch(int comparisonValue, int... elements){
return Arrays.stream(elements)
.anyMatch(e -> e == comparisonValue);
}
then call it like so:
if(anyMatch(10, a, b)){ ... }
That's not going to work like that. You're checking the value of two variables against a value, which ends up being two checks, if(a == 10 || b == 10).
However, you can modify this check to this code:
if(Arrays.asList(a,b).contains(10))
It results in the same behavior, but this is neither shorter nor easier to read.
Yeah turns out there isn't a way to make it shorter.
No, we can't do it because in case of java, there is no option for comparison of variables like that.
Even you couldn't write like this
if(a||b){ //staff }
but if you would write then you will get this error message
error: bad operand types for binary operator '||'
Not shorter, but more "intuitively" readable:
boolean condA = (a == 10);
boolean condB = (b == 10);
if(condA || condA){
//stuff
}
always keep in mind, the goal isn't to write shortest possible code, but best maintainable code.

Chained ANDs or chained ORs best practice

Which is better in terms of best practice / efficiency?
if (x == 1
&& y == 1
&& z == 1)
{ do things }
or
if (x != 1 ||
y != 1 ||
z != 1)
{ don't do things and go to a different bit of logic.}
Is there any difference in efficiency when short circuiting ANDs and ORs? Is it (generally) better to check positively or negatively when multiple logical assertions need to be made?
For pure optimization of the code it depends case-by-case. The scenario that will on average do the least amount of comparisons.
For code design it is also case-by-case. The if-cases should match what you are actually looking for. A function that tests if a string is inputted correctly for example. (the tests are made up)
public boolean isValidString (string s) {
if (s.isEmpty())
return false;
if (s.length() < 12)
return false;
if (s...)
return false
return true;
}
In this case the most logical approach is the ||. It could be written.
public boolean isValidString (string s) {
if (s.isEmpty() || s.length() < 12 || s...)
return false;
return true;
}
With http://en.wikipedia.org/wiki/De_Morgan%27s_laws this could be rewritten to not and. However it is not what we want to test, even though they yield the same result.
So stick to the logical approach in general cases.
If you think about efficiency then think about how often each case will occur. The most likely one should be put in front so the whole expression is shortcircuited immediately.
Better you use "==" instead of going for "!=".
This is also recommended with PMD.
The following is good and improves redability.
If(true){
//
}else{
//
}
than
If(!true){
//
}else{
//
}
Well, in some JVM implementations boolean values are stored as integers in the JVM. int value 1 meaning true and int value 0 meaning false. Also, comparison logic at processor level is architecture dependent. Some machines might subtract 2 operands, then add and then compare, others might compare byte by byte etc.. So, unless you are looking at a specific hardware architecture (which you shouldn't.. atleast for java programming language), I don't think this matters much..

Does Java waste time to check conditionA in "if(isOK && conditionA)" if isOK=false?

Ok, i am building program to check many fields. If at least 1 field is not ok then i don't want my program to spend time to check other fields. So let look at this code:
// Util.isReadyToUse method return true if the string is ready for using, & return false if it is not.
boolean isOK=true;
if(!Util.isReadyToUse(firstName)){
isOK=false;
}
else if(isOK && !Util.isReadyToUse(lastName)){
isOK=false;
}
else if(isOK && !Util.isReadyToUse(email)){
isOK=false;
}
.....more checking
if(isOK) {
//do sthing
}
Ok, when running, the program will first check !Util.isReadyToUse(firstName). Suppose it returns (isOK=false). Next the program will check isOK && !Util.isReadyToUse(lastName).
So my question here is that Since the isOK currently false, then will the program spend time to check the condition !Util.isReadyToUse(lastName) after &&?
Ok, As a human being, if you see isOK=false and now you see isOK && !Util.isReadyToUse(email), then you don't want to waste time to look at !Util.isReadyToUse(email) since isOK=false and u saw && after isOK.
Will machine also work like that?
I am thinking to use break but why people say break doesn't work in if statement:
if(!Util.isReadyToUse(firstName)){
isOK=false;
break;
}
else if(isOK && !Util.isReadyToUse(lastName)){
isOK=false;
break;
}......
What is the best solution in this situation?
So my question here is that Since the isOK currently false, then will
the program spend time to check the condition
!Util.isReadyToUse(lastName) after &&?
Java is smart, if you have a condition if(somethingFlase && something), then something won't be reached due to Short-circuit evaluation. Since the whole expression will be false regardless of the second condition, there is no need for Java to evaluate that.
From 15.23. Conditional-And Operator &&:
If the resulting value is false, the value of the conditional-and
expression is false and the right-hand operand expression is not
evaluated. If the value of the left-hand operand is true, then the right-hand expression is evaluated.
if(a && b) - if a is false, b won't be checked.
if(a && b) - if a is true, b will be checked, because if it's false, the expression will be false.
if(a || b) - if a is true, b won't be checked, because this is true anyway.
if(a || b) - if a is false, b will be checked, because if b is true then it'll be true.
No, it shortcuts the rest of the predicate.
That's you'll see things like
if(A != null && A.SomeVal == someOtherVal)
Java supports what is referred to as Short-Circuit Evaluation. See this page:
http://en.wikipedia.org/wiki/Short-circuit_evaluation
What this means is that if the first boolean in your statement is enough to satisfy the statement, then the rest of the values are skipped. If we have the following:
boolean a = false;
boolean b = true;
if(a && b) /*Do something*/;
'b' will never be checked, because the false value for 'a' was enough to break out of the if statement.
That being said, your program will never take advantage of this because the only time isOK is set to false is within one of your else if statements.
As the other responders mentioned Java will do the smart thing.
But it could be the case that you want Java to continue to check, in that case you can use & vs && or | vs ||.
if (someMethod() | anotherMethod() {
If the first method reutrns true, Java will still execute the second method.
if (someMethod() & anotherMethod() {
If the first method is false, Java will still execute the second method.
No, Java won't "waste time" for it. It's called short circuit evaluation.
This mechanism is commonly used e.g. for null checking :
if (foo != null && foo.neverFailsWithNPE()) {
// ...
}
You don't need to use break on an if..else if.. else statement because once it finds a condition which is true the rest aren't even looked at.

In Java, what are the boolean "order of operations"?

Let's take a simple example of an object Cat. I want to be sure the "not null" cat is either orange or grey.
if(cat != null && cat.getColor() == "orange" || cat.getColor() == "grey") {
//do stuff
}
I believe AND comes first, then the OR. I'm kinda fuzzy though, so here are my questions:
Can someone walk me through this statement so I'm sure I get what happens?
Also, what happens if I add parentheses; does that change the order of operations?
Will my order of operations change from language to language?
The Java Tutorials has a list illustrating operator precedence. The equality operators will be evaluated first, then &&, then ||. Parentheses will be evaluated before anything else, so adding them can change the order. This is usually pretty much the same from language to language, but it's always a good idea to double check.
It's the small variations in behavior that you're not expecting that can cause you to spend an entire day debugging, so it's a good idea to put the parentheses in place so you're sure what the order of evaluation will be.
Boolean order of operations (in all languages I believe):
parens
NOT
AND
OR
So your logic above is equivalent to:
(cat != null && cat.getColor() == "orange") || cat.getColor() == "grey"
The expression is basically identical to:
if ( (cat != null && cat.getColor() == "orange") || cat.getColor() == "grey") {
...
}
The order of precedence here is that AND (&&) has higher precedence than OR (||).
You should also know that using == to test for String equality will sometimes work in Java but it is not how you should do it. You should do:
if (cat != null && ("orange".equals(cat.getColor()) || "grey".equals(cat.getColor()))) {
...
}
ie use the equals() methods for String comparison, not == which simply does reference equality. Reference equality for strings can be misleading. For example:
String a = new String("hello");
String b = new String("hello");
System.out.println(a == b); // false
First, your if statement contains three main expressions:
cat != null
cat.getColor() == "orange"
cat.getColor() == "grey"
The first expression simply checks whether cat is not null. Its necessary otherwise the the second expression will get executed and will result in a NPE(null pointer excpetion). That's why the use of && between the first and second expression. When you use &&, if the first expression evaluates to false the second expression is never executed.
Finally you check whether the cat's color is grey.
Finally note that your if statement is
still wrong because if cat is
null, the third expression is still
executed and hence you get a null
pointer exception.
The right way of doing it is:
if(cat != null && (cat.getColor() == "orange" || cat.getColor() == "grey")) {
//do stuff
}
Check the order of parenthesis.
Yeah && is definitely evaluated before ||. But I see you are doing cat.getColor() == "orange" which might give you unexpected result. You may want to this instead :
if(cat != null && ("orange".equals(cat.getColor()) || "grey".equals(cat.getColor()))) {
//do stuff
}
Order of Operation is not what you need, you need boolean algebra, this includes boolean functions. Maxterms/minterms, Gray code, Karnaugh tables, diodes,transistors, logic gates, multiplexers, bitadders, flip flops...
What you want is to implement boolean "logic" on computers or virtual machines. With "order of operations" you may refer something about physics like managing delays on logic gates (OR, if) nanoseconds intervals?

First Java program (calculator) problems

I'm in the process of learning Java and my first project is a calculator, however I've run into a snag. I'm trying to get my calculator to let me enter a number then click an operator (+, -, x, /), enter another number then hit an operator again and have the display update and be able to keep this going.
Example, I would like to be able to hit the following and have it display the total each time I hit an operator after the first:
a + b / c - d =
The code I have seems (to me) like it should work but it doesn't. What am I doing wrong?
The following is the code I'm using when you hit an operator. By default wait is set to false. After running through the class once, value1 is stored and wait is set to true and that works fine. From there it doesn't seem to work quite right:
class OperatorListener implements ActionListener {
public void actionPerformed(ActionEvent event) {
String input = event.getActionCommand();
// Set display as string
String s = display.getText();
if (!wait) {
// Convert first input string to double
try {
value1 = Double.valueOf(s.trim()).doubleValue();
} catch (NumberFormatException nfe) {
System.out.println("NumberFormatException: " + nfe.getMessage());
}
dec = false;
} else {
// Convert second input string to double
try {
value2 = Double.valueOf(s.trim()).doubleValue();
} catch (NumberFormatException nfe) {
System.out.println("NumberFormatException: " + nfe.getMessage());
}
// Determine operation to be performed
if (operator == "add") {
value1 = Operators.add(value1, value2);
} else if (operator == "subtract") {
value1 = Operators.subtract(value1, value2);
} else if (operator == "multiply") {
value1 = Operators.multiply(value1, value2);
} else if (operator == "divide") {
value1 = Operators.divide(value1, value2);
}
// Convert final value to string and display
display.setText(Double.toString(value1));
dec = false;
}
// Determine operator hit
if (input.equals("+")) {
operator = "add";
} else if (input.equals("-")) {
operator = "subtract";
} else if (input.equals("x")) {
operator = "multiply";
} else if (input.equals("/")) {
operator = "divide";
}
// Set wait
wait = true;
}
}
EDIT: Updated code to fix some confusion and update the if statement. Even after this the same problem still exists. Also, the full source is available here
A few suggestions.
First, I would suggest when using a boolean as a conditional for an if statement, avoid comparison with true and false -- there are only two states for boolean anyway. Also, since there are only two states, rather than using else if (false), an else will suffice:
if (condition == true)
{
// when condition is true
}
else if (condition == false)
{
// when condition is false
}
can be rewritten as:
if (condition)
{
// when condition is true
}
else
{
// when condition is false
}
Second, rather than comparing the string literals "add", "subtract" and such, try to use constants (final variables), or enums. Doing a String comparison such as (operator == "add") is performing a check to see whether the string literal "add" and the operator variable are both refering to the same object, not whether the values are the same. So under certain circumstances, you may have the operator set to "add" but the comparison may not be true because the string literal is refering to a separate object. A simple workaround would be:
final String operatorAdd = "add";
// ...
if (input.equals("+"))
operator = operatorAdd;
// ...
if (operator == operatorAdd)
// ...
Now, both the assignment of operator and the comparison of operator both are referecing the constant operatorAdd, so the comparison can use a == rather than a equals() method.
Third, as this seems like the type of calculator which doesn't really require two operands (i.e. operand1 + operand2), but rather a single operand which is acting upon a stored value (i.e. operand + currentValue), it probably would be easier to have some variable that holds the current value, and another variable that holds the operator, and a method which will act according to the current operator and operand. (More or less an idea of an accumulator machine, or 1-operand computer.)
The basic method of operation will be:
Set the currentValue.
Set the operator.
Set the operand.
Perform the calculation.
Set the currentValue to the result of the calculation.
Set the operator to blank state.
Each step should check that the previous step took place -- be sure that an operation is specified (operator is set to a valid operator), then the next value entered becomes the operand. A calculator is like a state machine, where going from one step to another must be performed in a certain order, or else it will not proceed to the next step.
So, the calculator may be like this (pseudocode!):
// Initialize calculator (Step 1)
currentValue = 0;
operand = 0;
operator = operatorNone;
loop
{
operand = getOperand(); // Step 2
operator = getOperator(); // Step 3
// Step 4 and 5
if (operator == operatorAdd)
currentValue += operand;
if (operator == operatorSubtract)
currentValue -= operand;
// ...
// Step 6
operator = operatorNone;
}
Although the above code uses a single loop and doesn't work like a event-based GUI model, but it should outline the steps that it takes to run a calculator.
Whenever you enter an operator, your code will execute this:
Double.valueOf(s.trim())
for setting either value1 or value2 (depending on wait). This will throw an exception because operators can't be parsed as doubles. You might have better luck checking for the operator first, before trying to parse the input as a number. Then if it was an operator, you can skip the number parsing part.
Also consider what might happen if somebody were to enter two numbers or two operators in a row.
As Greg said, no matter what the input and no matter what the current program state, you always parse out number. You need to track the program state more cleanly. I assume that when you code has "String s = output.getText();" that you really mean "String s = input.getText();".
Also note that
if (wait == false) {
// Stuff for !wait
} else if (wait == true) {
// Stuff for wait
}
is unnecessarily redundant. You can replace it with:
if (!wait) {
// Stuff for !wait
} else {
// Stuff for wait
}
You should probably check the input string to see if it is an operator, first, and if it isn't then make sure it is numeric. Writing an infix calculator (that properly handles precedence) is not trivial.
After searching high and low I finally determined that the problem didn't lie within the code I provided. I had had a "wait = false;" in my NumberListener class that was screwing up the execution. To solve this I created 2 separate wait variables and all is working fine so far.
Thanks for the help and the tips guys, +1 to all of you for trying.
You could use the scripting engine in Java. If you don't have Java 6+, you can use Rhino which does the same thing. You can then do pretty much anything you can do in JavaScript
// create a script engine manager
ScriptEngineManager factory = new ScriptEngineManager();
// create a JavaScript engine
ScriptEngine engine = factory.getEngineByName("JavaScript");
// expose a, b, c, d
engine.put("a", 1);
engine.put("b", 8);
engine.put("c", 2);
engine.put("d", 3);
// evaluate JavaScript code from String
Number value = (Number) engine.eval("a + b / c * d");
System.out.println(value);
For more examples

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