Why doesn't Java have compound assignment versions of the conditional-and and conditional-or operators? (&&=, ||=)

问题

So for binary operators on booleans, Java has &, |, ^, && and ||.

Let's summarize what they do briefly here:

JLS 15.22.2 Boolean Logical Operators &, ^, and |
JLS 15.23 Conditional-And Operator &&
JLS 15.24 Conditional-Or Operator ||

For &, the result value is true if both operand values are true; otherwise, the result is false.

For |, the result value is false if both operand values are false; otherwise, the result is true.

For ^, the result value is true if the operand values are different; otherwise, the result is false.

The && operator is like & but evaluates its right-hand operand only if the value of its left-hand operand is true.

The || operator is like |, but evaluates its right-hand operand only if the value of its left-hand operand is false.

Now, among all 5, 3 of those have compound assignment versions, namely |=, &= and ^=. So my question is obvious: why doesn't Java provide &&= and ||= as well? I find that I need those more than I need &= and |=.

And I don't think that "because it's too long" is a good answer, because Java has >>>=. There must be a better reason for this omission.

From 15.26 Assignment Operators:

There are 12 assignment operators; [...] = *= /= %= += -= <<= >>= >>>= &= ^= |=

A comment was made that if &&= and ||= were implemented, then it would be the only operators that do not evaluate the right hand side first. I believe this notion that a compound assignment operator evaluates the right hand side first is a mistake.

From 15.26.2 Compound Assignment Operators:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

As proof, the following snippet throws a NullPointerException, not an ArrayIndexOutOfBoundsException.

    int[] a = null;
    int[] b = {};
    a[0] += b[-1];

回答1:

Reason

The operators &&= and ||= are not available on Java because for most of the developers these operators are:

error-prone
useless

Example for `&&=`

If Java allowed &&= operator, then that code:

bool isOk = true; //becomes false when at least a function returns false
isOK &&= f1();
isOK &&= f2(); //we may expect f2() is called whatever the f1() returned value

would be equivalent to:

bool isOk = true;
if (isOK) isOk = f1();
if (isOK) isOk = f2(); //f2() is called only when f1() returns true

This first code is error-prone because many developers would think f2() is always called whatever the f1() returned value. It is like bool isOk = f1() && f2(); where f2() is called only when f1() returns true.

If the developer wants f2() to be called only when f1() returns true, therefore the second code above is less error-prone.

Else &= is sufficient because the developer wants f2() to be always called:

Same example but for `&=`

bool isOk = true;
isOK &= f1();
isOK &= f2(); //f2() always called whatever the f1() returned value

Moreover, the JVM should run this above code as the following one:

bool isOk = true;
if (!f1())  isOk = false;
if (!f2())  isOk = false;  //f2() always called

Compare `&&` and `&` results

Are the results of operators && and & the same when applied on boolean values?

Let's check using the following Java code:

public class qalcdo {

    public static void main (String[] args) {
        test (true,  true);
        test (true,  false);
        test (false, false);
        test (false, true);
    }

    private static void test (boolean a, boolean b) {
        System.out.println (counter++ +  ") a=" + a + " and b=" + b);
        System.out.println ("a && b = " + (a && b));
        System.out.println ("a & b = "  + (a & b));
        System.out.println ("======================");
    }

    private static int counter = 1;
}

Output:

1) a=true and b=true
a && b = true
a & b = true
======================
2) a=true and b=false
a && b = false
a & b = false
======================
3) a=false and b=false
a && b = false
a & b = false
======================
4) a=false and b=true
a && b = false
a & b = false
======================

Therefore YES we can replace && by & for boolean values ;-)

So better use &= instead of &&=.

Same for `||=`

Same reasons as for &&=:
operator |= is less error-prone than ||=.

If a developer wants f2() not to be called when f1() returns true, then I advice the following alternatives:

// here a comment is required to explain that 
// f2() is not called when f1() returns false, and so on...
bool isOk = f1() || f2() || f3() || f4();

or:

// here the following comments are not required 
// (the code is enough understandable)
bool isOk = false;
if (!isOK) isOk = f1();
if (!isOK) isOk = f2(); //f2() is not called when f1() returns false
if (!isOK) isOk = f3(); //f3() is not called when f1() or f2() return false
if (!isOK) isOk = f4(); //f4() is not called when ...

回答2:

Probably because something like

x = false;
x &&= someComplexExpression();

looks like it ought to be assigning to x and evaluating someComplexExpression(), but the fact that the evaluation hinges on the value of x isn't apparent from the syntax.

Also because Java's syntax is based on C, and no one saw a pressing need to add those operators. You'd probably be better off with an if statement, anyway.

回答3:

It is this way in Java, because it is this way in C.

Now the question why it is so in C is because when & and && became different operators (sometime preceding C's descent from B), the &= variety of operators was simply overlooked.

But the second part of my answer does not have any sources to back it up.

回答4:

One of Java's original aims was to be "Simple, Object Oriented, and Familiar." As applied to this case, &= is familiar (C, C++ have it and familiar in this context meant familiar to someone who knows those two).

&&= would not be familiar, and it would not be simple, in the sense that the language designers were not looking to think of every operator they could add to the language, so less extra operators are simpler.

回答5:

Largely because Java syntax is based on C (or at least the C family), and in C all those assignment operators get compiled to arithmetic or bitwise assembly instructions on a single register. The assignment-operator version avoids temporaries and may have produced more efficient code on early non-optimising compilers. The logical operator (as they are termed in C) equivalents (&&= and ||=) don't have such an obvious correspondence to single assembly instructions; they usually expand to a test and branch sequence of instructions.

Interestingly, languages like ruby do have ||= and &&=.

Edit: terminology differs between Java and C

回答6:

For Boolean vars, && and || would use short circuit evaluation while & and | don't, so you would expect &&= and ||= to also use short circuit evaluation. There is a good use case for this. Especially if you are iterating over a loop, you want to be fast, efficient and terse.

Instead of writing

foreach(item in coll)
{
   bVal = bVal || fn(item); // not so elegant
}

I want to write

foreach(item in coll)
{
  bVal ||= fn(item);    // elegant
}

and know that once bVal is true, fn() will not be called for the remainder of the iterations.

回答7:

'&' and '&&' are not the same as '&&' is a short cut operation which will not do if the first operand is false while '&' will do it anyway (works with both number and boolean).

I do agree that it make more sense to exist but it is not that bad if it is not there. I guess it was not there because C does not have it.

Really can't think of why.

回答8:

It is allowed in Ruby.

If I were to guess, I would say that it is not frequently used so it wasn't implemented. Another explanation could be that the parser only looks at the character before the =

回答9:

I cannot think of any better reason then 'It looks incredible ugly!'

回答10:

verifies both operands, it's a bitwise operator. Java defines several bitwise operators, which can be applied to the integer types, long, int, short, char, and byte.

stops evaluating if the first operand evaluates to false since the result will be false, it's a logical operator. It can be applied to booleans.

The && operator is similar to the & operator, but can make your code a bit more efficient. Because both expressions compared by the & operator must be true for the entire expression to be true, there’s no reason to evaluate the second expression if the first one returns false. The & operator always evaluates both expressions. The && operator evaluates the second expression only if the first expression is true.

Having a &&= assignment operator wouldn't really add new functionality to the language. The bitwise operator's arithmetic is much more expressive, you can do integer bitwise arithmetic, which includes Boolean arithmetic. The logical operators can merely do Boolean arithmetic.

回答11:

a&b and a && b are not the same thing.

a && b is a boolean expression that returns a boolean, and a & b is a bitwise expression that returns an int (if a and b are ints).

whare do you think they are the same?

来源：https://stackoverflow.com/questions/2324549/why-doesnt-java-have-compound-assignment-versions-of-the-conditional-and-and-co

标签

java

conditional-operator

assignment-operator

short-circuiting

compound-assignment