Doubt about statements in C / C ++ / Java

Question

Consider the following snippet:

int a,b; a = 1; b = 2; c = a++++b; // does not work!! Compilation error. c = a++*+b; // works !! 

Please help me understand this behavior.

Answer 1

 c = a++++b; 

regarded as:

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

which is wrong as you try to increase non-lvalue.

and

 c = a++*+b; 

regarded as:

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

The reason for this behavior is: the lexical analyzer of the C language is greedy .

In case 1: after the token 'a' (identifier), lexer sees + followed by the other +, so it consumes both (as an increment operator) as part of the same token. It does not make the third + part of the same token as +++ is not a valid token. Similarly, it groups the following two + into + tokens, making it the same as:

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

which is wrong, since ++ does not return an lvalue, so you cannot apply ++ to it. Something like saying 5 ++;

But in case 2: the first pair ++ will be grouped (as an increment operator). Then only one * will be a token, since you cannot combine it with + as * +, it is not a valid token. Finally, + will be a token (like unary +), effectively making your statement as follows:

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

You can override this greedy lexer behavior using parentheses or spaces as follows:

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

Answer 2

This is effective because of the " maximum munch rule " in C.

 c = a++++b; 

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

 c = a++*+b; 

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

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

If the input stream is parsed in the preprocessing token to the specified character, the next preprocessing token is the longest sequence of characters that can make up the preprocessing token.

Answer 3

For the same reason why we get an error in C ++ for:

 vector<vector<int>>; 

>> will be considered as one operator.

Answer 4

operator priority . ++ has a higher priority than binary +.

Answer 5

limit ++ is equal to priority +. Therefore, we use from left to right.