CODEWITHSUNDEEP
javareturn-valueincrement
gstackoverflow
gstackoverflow

Reputation: 37034

Why cannot I increment returned value?

I noticed a surprising behaviour.

I can write so:

Integer integer = 0;
integer++;

But I cannot write so:

getInteger()++;

Here's my getInteger() method:

public static Integer getInteger(){
    return 0;
}

For me, two snippets look the same.

But the second piece of code returns an error:

unexpected type
        getInteger()++;
                  ^
  required: variable
  found:    value

Why is the second construction forbidden?

Often, I feel forced to write "ugly" code like this:

obj.set(obj.get()+1);

It looks redundant.

Upvotes: 1

Views: 893

Answers (5)

Alnitak
Alnitak

Reputation: 339816

You are not returning an int; you're returning an Integer object. The Integer class is immutable and has no ++ operator of its own.

Also, the ++ operator requires that its operand be an lvalue, i.e. an assignable variable as opposed to a value. The result of getInteger() is the latter.

In the case of your first test:

Integer integer = 0; integer++;

Java's autoboxing allows the use of the ++ operator, but only because Java automatically converts the Integer into an int and then back again, and because the variable integer is an lvalue. It's more or less equivalent to:

int tmp_int = integer.intValue();
int result = tmp_int++;
integer = Integer.getInteger(tmp_int);
(void)result;

result - the effective "value" of the expression isn't integer because you've used the "post increment" version of ++, so the result is the value that integer had before it was incremented.

Upvotes: 0

Kartik_Koro
Kartik_Koro

Reputation: 1287

As it says, it requires a variable for ++ operator. ++ increments a value that is stored somewhere, so the incremented value is now stored in that location. Your function just returns a value, which is not a variable.

It's as simple as Rahul mentioned in the comment.

//This works
x=5;
x++; //Now x=6 since x++ is equivalent to x=x+1;

// This doesn't work
5++; // This doesn't even make sense, trying to do 5=5+1??

// Similarly, getNumber() is just a value that is returned, it is a value, not a variable
// So it is like the second case
// This doesn't work
getNumber()++; //Trying to do getNumber()=getNumber()+1; ??

Upvotes: 1

nobalG
nobalG

Reputation: 4620

8++ will not make sense, while

int x=8;
x++;

does make sense, because x++, is working like x+=1, i.e., x=x+1 and 8=8+1 is something not acceptable.

Upvotes: 2

Peter Lawrey
Peter Lawrey

Reputation: 533492

Why is the second construction forbidden?

Integer is immutable, so you couldn't change it anyway. int is mutable but since it only returns a copy on the value, this won't work either.

E.g.,

public static Integer getInteger(){
    return 0;
}

// what would this do
getInteger()++;

// now the method has been changed to 
public static Integer getInteger(){
    return 0+1;
}
// so now
System.out.println(getInteger()); // prints 1 ??

obj.set(obj.get()+1);

This breaks encapsulation as well. I would suggest a method like this:

AtomicInteger i = new AtomicInteger(0);
i.incrementAndGet();

Or you could have your own:

i.incr();

Upvotes: 0

Fred Foo
Fred Foo

Reputation: 363547

x++ is not the same as x + 1. It's short for x += 1, i.e. x = x + 1; increment, then assign. So getInteger()++ means getInteger() = getInteger() + 1, which makes no sense. You're adding one to a value, not incrementing a variable.

Apart from that, ++ works on int but not Integer, but even if you were returning an int you couldn't increment it with ++:

class Test {
    static public void main() {
        getint()++;
    }
    static int getint() {
        int a = 0;
        return a;
    }
}

gives

Test.java:3: error: unexpected type
        getint()++;
              ^
  required: variable
  found:    value

(In C++ you could make this work with references, but Java doesn't have those.)

Upvotes: 10

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