Why doesn't a+++++b work?

Question

int main ()
{
   int a = 5,b = 2;
   printf(\"%d\",a+++++b);
   return 0;
}

This code gives the following error:

error: lval

Answer 1

The lexer uses what's generally called a "maximum munch" algorithm to create tokens. That means as it's reading characters in, it keeps reading characters until it encounters something that can't be part of the same token as what it already has (e.g., if it's been reading digits so what it has is a number, if it encounters an A, it knows that can't be part of the number. so it stops and leaves the A in the input buffer to use as the beginning of the next token). It then returns that token to the parser.

In this case, that means +++++ gets lexed as a ++ ++ + b. Since the first post-increment yields an rvalue, the second can't be applied to it, and the compiler gives an error.

Just FWIW, in C++ you can overload operator++ to yield an lvalue, which allows this to work. For example:

struct bad_code { 
    bad_code &operator++(int) { 
        return *this;
    }
    int operator+(bad_code const &other) { 
        return 1;
    }
};

int main() { 
    bad_code a, b;

    int c = a+++++b;
    return 0;
}

The compiles and runs (though it does nothing) with the C++ compilers I have handy (VC++, g++, Comeau).