Over the years I've encountered some great topics regarding C++ that I found interesting enough to archive in my 'box o stuff'.
As I have nothing better to do on a Wednesday night, I'll convey this information via interpretive dance in puzzle form for anybody else looking for a time passer.If you know the answer hold off a little ( or for a few posts between Q & A ) for people to have a crack. If people like these sorts of things, I'll post more.
And before anyone asks, I'm not trying to get someone to do my homework.
#1 Increment This
For the first entry, here is my own extension to a common puzzle. This requires a short preface.
Take the code below, what is printed to the serial monitor? ( not the question )
int i = 0;
int j = 1;
Serial.println( i+++j, DEC );
For those of you who haven't seen this, it is a classic example of 'order of evaluation', without looking at associativity, operator precedence defines how this is evaluated.
Precedence
Operator
1
++ ( Suffix increment )
2
++ ( Prefix increment )
2
+ ( Unary plus )
So...
The suffix operator has precedence, and the code is:
i++ + j; //( i Suffix increment ) + j
rather than:
i + ++j; //i + ( j prefix increment )
As you can see operator precedence explains why this works... Or does it.
Why does the code below fail to compile while the two equivalent versions beneath it compile fine?
i+++++j;
i++ + ++j;
i+++ ++j;
Hint #1: the answer is not related to the spaces in the code, if you believe this is the answer, explain why the spaces didn't cause error with the version: 'i+++j'
I'll bump it back up for others to see anyway. This may be difficult, but I thought it was a good one for a first, the next one I have planned should be a little easier.
Hint #2 fails to compile because of the way C combines characters into diagraphs to create a lexical unit and operator precedence. The compiler sees this...
i++++j;
...as...
( ( i ++ ) ++ ) j ;
From the perspective of the outer operator (the second ++), the expression is...
( ( i + 1 ) ++ ) j ;
...which fails to compile because the increment operator requires a variable (an "lvalue") as the operand.
The five pluses is just an extension of the above...
C or C++ hasn't even come into play when this occurs, this happens even before macros are expanded.
The tokenizer creates the largest possible token from the source input, so the token '++' is considered before '+'.