Thread: Never encounter such coding

Hi all,

i have never encounter a code that is written this way. From my experience, isn't a code is suppose to end when ; is encounter? So why is the following code has a \ after ; . can someone explain to me what the following code means?

Code:

#define a   (b |= ~c); \          //What is \ after ;
            (d |= e); \
            (f &= ~g); \
            (h |= i);

Thanks

It's just a line continuation.
#defines can only exist on one line, so it's just a nice way of saying

#define a (b |= ~c); (d |= e); (f &= ~g); (h |= i);

your code tells a preprocessor (a specialised text editor) to replace a in your source code with:

Code:

(b |= ~c);    
(d |= e); 
(f &= ~g);
(h |= i);

before it is actually compiled.

a #define ... terminates at the end of line. putting \ there continues that line (to the one below).

that bit of code (#define a ...) is not actually c, which you are lead to believe. i've read that the syntax of the preprocessor is completely different to c.

I see, i see....well that should clear all the clouds...Thanks guys (or gals)...