Post Increment with substitution

Question

I have a question here already of my own but I want to extend it Post increment with example

char a = \'D\';
int b = 5;
System.out.println(a++/b+--a*b++);


        

Answer 1

Don't confused evaluation order with precedence.

Evaluation order states that operands are always executed left to right.

Precedence states that * and / operators are applied before + and -, unless overridden by parenthesis.

So, you're right that a++/b+--a*b++ means (a++ / b) + (--a * b++). That's precedence.

Since numeric operators promote values to int (in this case), you're also right that char a = 'D' is equivalent to int a = 68.

So:

(a++ / b) + (--a * b++)    a = 68   b = 5
(68  / b) + (--a * b++)    a = 69   b = 5
(68  / 5) + (--a * b++)    a = 69   b = 5
(68  / 5) + (68  * b++)    a = 68   b = 5
(68  / 5) + (68  * 5  )    a = 68   b = 6
13        + 340            a = 68   b = 6
353                        a = 68   b = 6

As you can see, b++ does have an effect: On the value of b after executing the expression.

Answer 2

Step 4: (68/6+68*5) its your wrong! b is 5 in all positions in the calculation instead of it's value 6.

so we have:

Step 4: (68/5+68*5) = (13+340)=353