java - Incompatible types. Required: short, Found: int

I just can't understand the difference between this:

short d = 0;
//some code

node.accessible = d + 1;

and this

short d = 0
//same code here
node.accessible = d;
node.accessible += 1;

the second thing is working, but the 1st one is't inteliji showes "incompatiable types" error.

p.s. node class:

public class Node {
int n;
short accessible;
Node(int n){
    this.n = n;
    this.accessible = -1;
}

}   

Answer

In the first version :

node.accessible = d + 1;

d + 1 produces a int as summing an int and a short produces an int.
The JLS states indeed that (look at the last case, emphasis is mine) :

5.6.2. Binary Numeric Promotion

When an operator applies binary numeric promotion to a pair of
operands, each of which must denote a value that is convertible to a
numeric type, the following rules apply, in order:

1. If any operand is of a reference type, it is subjected to unboxing
   conversion (§5.1.8).





2. Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:

   
   
   
   
   

   • If either operand is of type double, the other is converted to double.

   
   

   • Otherwise, if either operand is of type float, the other is converted
     to float.

   
   

   • Otherwise, if either operand is of type long, the other is converted
     to long.

   
   

   • Otherwise, both operands are converted to type int.

   
   

But you cannot assign a int to the accessible field that is a short without explicit cast as int has a broader range than short.

---

While in the second version, a Compound Assignment Operators is used (+=):

node.accessible += 1;

As a consequence, in your case the result of the operation is converted to short : the type of the left-hand variable as the JLS states :

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 Evaluations only once.

And more specifically in your case :

Otherwise, the result of the binary operation is converted to the type
of the left-hand variable, subjected to value set conversion (§5.1.13)
to the appropriate standard value set (not an extended-exponent value
set), and the result of the conversion is stored into the variable.