Why do I get an error in this program?
If I change the middle operator to minus like this (a++-++b), it is executing without any error:
class Demo
{
public static void main(String []args){
int a=10;
int b=20;
System.out.println (a+++++b);
}
}
The parser is unable to see what you want it to see. You want it to see
(a+++++b)
as
(a++ + ++b)
but, the parser is seeing it as
(a++++ +b)
and therefore, it is throwing a compilation error as ++ operator is expecting a variable but it is getting a++, which is a value.
The subtract sign works because it is identified as a separate operator from the series of additions. The additions however appear to become one string of operators. if you just add a space between the addition signs it should work fine as it will identify the addition operator separately.. eg System.out.println(a++ + ++b);
public class Demo {
public static void main(String[] args) {
int a=10;
int b=20;
System.out.println(a++ + ++b);
}
}
// output
31
I think you might want this, don't use +++ it's not valid operator in java, BTW, you should post better question next time.
For
(a++-++b);
The parser knows, that there is no such thing as an +- or -+ operator on int. So it can split the expression in (a++ - ++b);
For
(a+++++b);
This can be (a+ ++... or (a++ +... but let's first have a look at a simpler expression:
int a = 100;
int b = 0;
-> int c=a+++b;
| Added variable c of type int with initial value 100
What is the desired result? a++ +b or a+ ++b? Well, it's interpreted as c= (a++)+b; but it's bad style, since you have to have the details of operator precedence in head.
Is ++++b allowed, or a++++?
-> ++++b;
| Error:
| unexpected type
| required: variable
| found: value
| ++++b;
| ^-^
-> a++++;
| Error:
| unexpected type
| required: variable
| found: value
| a++++;
| ^-^
No. Why not? You could argue, that it is (a++)++; or ++(++b); But both include two steps, increment by one and returning a value, so for
a=100
b=0
c=a++;
c=++b;
we know, that c will become 100 or 1, in one case, the return is performed before incrementing (c=a++), in the other case after (++b);
Now what should happen with ++++b; or ++(++b); What get's returned?
Considering that, the only valid interpretation for a+++++b is a++ + ++b; There is no a+ ++ ++b, nor a+ +++ +b.
And a++ + ++b; works:
-> c=a++ + ++b;
| Variable c has been assigned the value 101
But since it is very rarely used, and not a big deal to insert two spaces, and more reader friendly, anyhow, we should live with the inconsistency, of a+++++b being rejected, while a++-++b is allowed, which doesn't make the latter a nice to read expression. Horizontal spaces make code better parsable for humans too, and that should be your main concern.
If you like to drive someone crazy, use
c=a--- - --b;
since there is this unary -, which is handy for producing negative values like -7.
Here, as + operator and ++ operator have equal precedence (similarly, - and -- have equal precedence.) So, +++ and --- gives an error as it will be broken into '+' and '++' ('-' and '--').
To solve this error, we can use +(++).
Hope that helps.
a+++++b ,a++---b,a------b and a--+++b will throw an error because + and ++ have same precedence.So Java will throw invalid argument for +++b,---b,---b and +++b respectively.
