Java – Understanding the =+ Operator

javamathoperators

Consider this code:

long val = 0;
for(int i = 0; i < 2; val++)
    val =+ ++i;

System.out.println(val);

Why is val = 3 in the end?

I would have calculated like this:

val     i
0       0   i < 2 = true;
0       0   ++i;
0       1   val =+ 1;
1       1   (end of for loop) val++;
2       1   i < 2 = true;
2       1   ++i;
2       2   val =+ 2;
4       2   (end of for loop) val++;
5       2   i < 2 = false;
Output: 5

But it's 3. I don't understand why the increment val =+ ++i is not done the second time when i = 1 and getting pre-incremented to i = 2.

Best Answer

Let's focus on the unusual-looking line first:

val =+ ++i;

The operators here are = (assignment), + (unary plus), and ++ (pre-increment). There is no =+ operator. Java interprets it as two operators: = and +. It's clearer with appropriate whitespace added:

val = + ++i;

Now let's analyze the processing:

First iteration: val and i are 0. i is pre-incremented to 1, and that's the result of ++i. The unary + does nothing, and 1 is assigned to val. Then the iteration statement val++ occurs and now val is 2. i is still 1, so the for loop condition is met and a second iteration occurs.

Second iteration: i is pre-incremented again, to 2. The unary + does nothing and val is assigned 2. The iteration statement val++ occurs again and it's now 3. But i is now 2, and it's not less than 2, so the for loop terminates, and val -- 3- is printed.

Related Solutions

Java – Difference Between += and =+ Operators

a += b is short-hand for a = a + b (though note that the expression a will only be evaluated once.)

a =+ b is a = (+b), i.e. assigning the unary + of b to a.

Examples:

int a = 15;
int b = -5;

a += b; // a is now 10
a =+ b; // a is now -5

Java – Understanding the ‘+=’ Operator

The "common knowledge" of programming is that x += y is an equivalent shorthand notation of x = x + y. As long as x and y are of the same type (for example, both are ints), you may consider the two statements equivalent.

However, in Java, x += y is not identical to x = x + y in general.

If x and y are of different types, the behavior of the two statements differs due to the rules of the language. For example, let's have x == 0 (int) and y == 1.1 (double):

    int x = 0;
    x += 1.1;    // just fine; hidden cast, x == 1 after assignment
    x = x + 1.1; // won't compile! 'cannot convert from double to int'

+= performs an implicit cast, whereas for + you need to explicitly cast the second operand, otherwise you'd get a compiler error.

Quote from Joshua Bloch's Java Puzzlers:

(...) compound assignment expressions automatically cast the result of the computation they perform to the type of the variable on their left-hand side. If the type of the result is identical to the type of the variable, the cast has no effect. If, however, the type of the result is wider than that of the variable, the compound assignment operator performs a silent narrowing primitive conversion [JLS 5.1.3].

Best Answer

Related Solutions

Java – Difference Between += and =+ Operators

Java – Understanding the ‘+=’ Operator

Related Question