Operators in C/C++/Java

Question

Consider the following fragment:

int a,b;
a = 1;
b = 2;

c = a++++b; // does not work!! Compilation error.
c = a++*+b; // works !!

Help me understand this behaviour.

Answer 1

c = a++++b; 

is treated as:

c = ((a++)++)b;  

which is incorrect as you are trying to increment non-lvalue.

and

c = a++*+b; 

is treated as:

c = (a++)*(+b);

The cause for this behaviour is: The C language lexical analyzer is greedy.

In case 1: After the token 'a' (identifier) the lexer sees +, followed by another +, so it consumes both (as the increment operator) as part of same token. It does not make the 3rd + part of the same token as +++ is not a valid token. Similarly it groups the next two + into ++ token making it effectively same as:

c = ((a++)++)b;

which is not correct as a++ will not return a lvalue, hence you can't apply a ++ on it. Something similar to saying 5++;

But in case2: the first pair of ++ will be grouped together (as increment operator). Next the * alone will be a token as you cannot combine it with a + as *+ is not a valid token. Finally the + will a token (as unary +) effectively making your statement as:

c = (a++)*(+b);

You can override this greedy behaviour of the lexer by making use of parenthesis or whitespaces as follows:

c = a++ + +b;  
c = a++ * +b;  

Answer 2

This is effectively because of "maximum munch rule" in C.

c = a++++b;

is parsed as c = a++ ++ b;, which is syntax error.

c = a++*+b;

is parsed as c = a++ * +b;, which is OK.

From the C99 draft, section 6.4p4 (emphasis mine):

If the input stream has been parsed into preprocessing tokens up to a given character, the next preprocessing token is the longest sequence of characters that could constitute a preprocessing token.