Why this boolean statement is true?
a= 10;
b = 0;
7 < a || a == b && b > 9 - a / b
Since anything divided by 0 is error
Since the first operand of the OR (||) operator (a > 7) evaluates to true, it short circuits and nothing else is evaluated. Therefore the entire expression evaluates to true.
7 < a returns true. Since it's a || after, the rest isn't executed.
This is because true || false is true, and true || true is true too, so evaluing the second member is but a waste of time.
Your OR-Operator || uses lazy evaluation or short-circuit evaluation. This means, since the very first expression 7 < ais true, it won't evaluate any other statements including the one with a division by zero, since java already found something true.
If you actually want to get an error, you can use this OR-Operator | which should enforce the evaluation of all statements. Most only use it as a bitwise-operator, but its also a non-short-circuit version of ||. For a more in-depth look at || vs. |, look here.
For example,
boolean c = (7 < a | a == b && b > 9 - a / b);
will cause an ArithmeticExcption, as expected.
I have a question about IF clause in Java.
I have and expression:
if ((someObject != null & connectedToTheInternet) || operate) {
// some action
}
Is my logic right: if someObject != null equals to true and connectedToTheInternet equals false then we have (someObject != null & connectedToTheInternet) equals false and then we have the following block:
if (false || operate) {
// some action
}
And if operate equals true then // some action will be triggered?
Just a first note: logical AND operator is "&&" and not just "&" (bitwise AND).
Following, your answer is YES... JVM will run conditions following your thinking.
By the way, I suggest you to read something abot short-circuit of these operators: it can be interesting if you are learning.
For example if you have if (a && (b || c) and a==false, then JVM won't evaluate b or c conditions because false && ... will be always false.
The same in the case of if (a || b || c && d): if a==true then JVM will ignore the other parts and consider it as true because true || .... will be always true.
Yes. if-clauses are evaluated from left to right. If no parenthesis are used, && has precedence, even higher precedence has ! (not) - similar to multiplication (AND), addition (OR) and negative numbers (-) in math (e.g. "A && (B || C) != A && B || C == (A && B) ||C") - I recommend to use parenthesis if you are unsure. This makes it possible to combine != null checks and calls to methods in the same if statement (e.g., if (object!=null && object.dosomething())).
Btw. there is a difference between & and && (short-circuit), when & is used, the second condition gets evaluated even if the first is false already. When && is used, Java doesn't check the second condition if the first one is false as the whole term cannot be true anymore - this is only important when the second condition is not a boolean variable but a method (which gets called or not -> side effects; this "skipping" is called short-circuit). Same for || and | where the second operator might not get evaluated if the first is already true (in case of ||).
Normally only ||and && are used.
Just for completeness: & is also the bitwise AND operator in Java.
if (false || operate) {
// some action
}
If operate true then if block executing.
This question already has answers here:
Java logical operator short-circuiting
(10 answers)
Closed 8 years ago.
I am not convinced why below statement is valid and does not throw exception.
ArrayList <String> some_list = null;
if (some_list != null && some_list.size() > 0) {
// do something...not important if I get here or not.
}
some_list is null, and first check in if() is valid, but what about getting size on a null reference?
isn't this illegal ?
or does it work this way:
if (codition1 && condition2) {
}
only if condition1 is true check condition2 ?
since I am not convinced , although I confirmed by writing test code. I always do like below:
if (some_list != null) {
if (some_list.size() > 0) {
}
}
Please help me understand the logical && and null point checking in if statement.
Thank you.
&& is "short circuiting". The expression on the right is never executed if the expression on the left evaluates to false.
The same is true for ||, which never executes the expression on the right if the expression on the left is true.
& and | are normally used for bit operations on integers, but can also be used on booleans. These are not short-circuiting. If you had done
if ((some_list != null) & (some_list.size() > 0)) {
then it would have failed in exactly the way you're asking about.
This question already has answers here:
What is the difference between & and && in Java?
(15 answers)
Closed 8 years ago.
Case 1:
boolean t=true;
boolean f= false,b;
b=(t && ((i++)==0));
b=(f && ((i+=2)>0));
System.out.println(i); // prints i=1
Case 2:
boolean t=true;
boolean f= false,b;
b=(t & ((i++)==0));
b=(f & ((i+=2)>0));
System.out.println(i); // prints i=3
case 1 :
(f && ((i+=2)>0 --> f is false, so second statement will not be executed.
case 2 :you are not using binary && , you are using bitwise &, so, i=i+2 will be executed. So result is 3.
This is because when you use logical and (&&) and the first operand is false, then the other operand is not evaluated.
b=(f && ((i+=2)>0));
here the (i+=2) never happens.
When you do a bitwise and (&), both operands are evaluated. (i+=2) is executed, hence the different result. To learn more about operators in Java, read: http://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html
& = arithmetic AND
&& = logical AND
in this code
`b=(t & ((i++)==0));
b=(f & ((i+=2)>0));`
i is incremented
In the code
b=(f && ((i+=2)>0));
a "short circuit" is preformed.
f is false for so the rest of the equation does not need to be calculated so i is not increased by 2.
I know the rules for && and || but what are & and |? Please explain these to me with an example.
Those are the bitwise AND and bitwise OR operators.
int a = 6; // 110
int b = 4; // 100
// Bitwise AND
int c = a & b;
// 110
// & 100
// -----
// 100
// Bitwise OR
int d = a | b;
// 110
// | 100
// -----
// 110
System.out.println(c); // 4
System.out.println(d); // 6
Thanks to Carlos for pointing out the appropriate section in the Java Language Spec (15.22.1, 15.22.2) regarding the different behaviors of the operator based on its inputs.
Indeed when both inputs are boolean, the operators are considered the Boolean Logical Operators and behave similar to the Conditional-And (&&) and Conditional-Or (||) operators except for the fact that they don't short-circuit so while the following is safe:
if((a != null) && (a.something == 3)){
}
This is not:
if((a != null) & (a.something == 3)){
}
"Short-circuiting" means the operator does not necessarily examine all conditions. In the above examples, && will examine the second condition only when a is not null (otherwise the whole statement will return false, and it would be moot to examine following conditions anyway), so the statement of a.something will not raise an exception, or is considered "safe."
The & operator always examines every condition in the clause, so in the examples above, a.something may be evaluated when a is in fact a null value, raising an exception.
I think you're talking about the logical meaning of both operators, here you have a table-resume:
boolean a, b;
Operation Meaning Note
--------- ------- ----
a && b logical AND short-circuiting
a || b logical OR short-circuiting
a & b boolean logical AND not short-circuiting
a | b boolean logical OR not short-circuiting
a ^ b boolean logical exclusive OR
!a logical NOT
short-circuiting (x != 0) && (1/x > 1) SAFE
not short-circuiting (x != 0) & (1/x > 1) NOT SAFE
Short-circuit evaluation, minimal evaluation, or McCarthy evaluation (after John McCarthy) is the semantics of some Boolean operators in some programming languages in which the second argument is executed or evaluated only if the first argument does not suffice to determine the value of the expression: when the first argument of the AND function evaluates to false, the overall value must be false; and when the first argument of the OR function evaluates to true, the overall value must be true.
Not Safe means the operator always examines every condition in the clause, so in the examples above, 1/x may be evaluated when the x is, in fact, a 0 value, raising an exception.
I know there's a lot of answers here, but they all seem a bit confusing. So after doing some research from the Java oracle study guide, I've come up with three different scenarios of when to use && or &.
The three scenarios are logical AND, bitwise AND, and boolean AND.
Logical AND:
Logical AND (aka Conditional AND) uses the && operator. It's short-circuited meaning: if the left operand is false, then the right operand will not be evaluated. Example:
int x = 0;
if (false && (1 == ++x) {
System.out.println("Inside of if");
}
System.out.println(x); // "0"
In the above example the value printed to the console of x will be 0, because the first operand in the if statement is false, hence java has no need to compute (1 == ++x) therefore x will not be computed.
Bitwise AND:
Bitwise AND uses the & operator. It's used to preform a bitwise operation on the value. It's much easier to see what's going on by looking at operation on binary numbers ex:
int a = 5; // 5 in binary is 0101
int b = 12; // 12 in binary is 1100
int c = a & b; // bitwise & preformed on a and b is 0100 which is 4
As you can see in the example, when the binary representations of the numbers 5 and 12 are lined up, then a bitwise AND preformed will only produce a binary number where the same digit in both numbers have a 1. Hence 0101 & 1100 == 0100. Which in decimal is 5 & 12 == 4.
Boolean AND:
Now the boolean AND operator behaves similarly and differently to both the bitwise AND and logical AND. I like to think of it as preforming a bitwise AND between two boolean values (or bits), therefore it uses & operator. The boolean values can be the result of a logical expression too.
It returns either a true or false value, much like the logical AND, but unlike the logical AND it is not short-circuited. The reason being, is that for it to preform that bitwise AND, it must know the value of both left and right operands. Here's an ex:
int x = 0;
if (false & (1 == ++x) {
System.out.println("Inside of if");
}
System.out.println(x); //"1"
Now when that if statement is ran, the expression (1 == ++x) will be executed, even though the left operand is false. Hence the value printed out for x will be 1 because it got incremented.
This also applies to Logical OR (||), bitwise OR (|), and boolean OR (|)
Hope this clears up some confusion.
The operators && and || are short-circuiting, meaning they will not evaluate their right-hand expression if the value of the left-hand expression is enough to determine the result.
& and | provide the same outcome as the && and || operators. The difference is that they always evaluate both sides of the expression where as && and || stop evaluating if the first condition is enough to determine the outcome.
In Java, the single operators &, |, ^, ! depend on the operands. If both operands are ints, then a bitwise operation is performed. If both are booleans, a "logical" operation is performed.
If both operands mismatch, a compile time error is thrown.
The double operators &&, || behave similarly to their single counterparts, but both operands must be conditional expressions, for example:
if (( a < 0 ) && ( b < 0 )) { ... } or similarly,
if (( a < 0 ) || ( b < 0 )) { ... }
source: java programming lang 4th ed
& and | are bitwise operators on integral types (e.g. int): http://download.oracle.com/javase/tutorial/java/nutsandbolts/op3.html
&& and || operate on booleans only (and short-circuit, as other answers have already said).
Maybe it can be useful to know that the bitwise AND and bitwise OR operators are always evaluated before conditional AND and conditional OR used in the same expression.
if ( (1>2) && (2>1) | true) // false!
&& ; || are logical operators.... short circuit
& ; | are boolean logical operators.... Non-short circuit
Moving to differences in execution on expressions. Bitwise operators evaluate both sides irrespective of the result of left hand side. But in the case of evaluating expressions with logical operators, the evaluation of the right hand expression is dependent on the left hand condition.
For Example:
int i = 25;
int j = 25;
if(i++ < 0 && j++ > 0)
System.out.println("OK");
System.out.printf("i = %d ; j = %d",i,j);
This will print i=26 ; j=25, As the first condition is false the right hand condition is bypassed as the result is false anyways irrespective of the right hand side condition.(short circuit)
int i = 25;
int j = 25;
if(i++ < 0 & j++ > 0)
System.out.println("OK");
System.out.printf("i = %d ; j = %d",i,j);
But, this will print i=26; j=26,
If an expression involving the Boolean & operator is evaluated, both operands are evaluated. Then the & operator is applied to the operand.
When an expression involving the && operator is evaluated, the first operand is evaluated. If the first operand evaluates to false, the evaluation of the second operand is skipped.
If the first operand returns a value of true then the second operand is evaluated. If the second operand returns a value of true then && operator is then applied to the first and second operands.
Similar for | and ||.
While the basic difference is that & is used for bitwise operations mostly on long, int or byte where it can be used for kind of a mask, the results can differ even if you use it instead of logical &&.
The difference is more noticeable in some scenarios:
Evaluating some of the expressions is time consuming
Evaluating one of the expression can be done only if the previous one was true
The expressions have some side-effect (intended or not)
First point is quite straightforward, it causes no bugs, but it takes more time. If you have several different checks in one conditional statements, put those that are either cheaper or more likely to fail to the left.
For second point, see this example:
if ((a != null) & (a.isEmpty()))
This fails for null, as evaluating the second expression produces a NullPointerException. Logical operator && is lazy, if left operand is false, the result is false no matter what right operand is.
Example for the third point -- let's say we have an app that uses DB without any triggers or cascades. Before we remove a Building object, we must change a Department object's building to another one. Let's also say the operation status is returned as a boolean (true = success). Then:
if (departmentDao.update(department, newBuilding) & buildingDao.remove(building))
This evaluates both expressions and thus performs building removal even if the department update failed for some reason. With &&, it works as intended and it stops after first failure.
As for a || b, it is equivalent of !(!a && !b), it stops if a is true, no more explanation needed.