[prog] C code

[Kathryn Hogg]

Noir said:
> Can anyone help me understand the following C code?
> Like, why the output is 4.
>
> #include <stdio.h>
> #define SIX    1 + 1
> #define NINE   2 + 1
> int main(void)
>  {
>     printf( "What you get if you multiply six by nine:
> %d\n",
>                 SIX * NINE
>                           );
>      return 0;
> }

SIX * NINE is expanded by the preprocessor as

1 + 1 * 2 + 1

* has a higher precedence than + so that evaluates as

1 + (1 * 2) + 1

You could do your defines as
#define SIX  (1+1)
#define NINE (2+1)

which would make your expression evaluate to (1+1) * (2+1)  which is 6.

-- 
Kathryn
http://womensfooty.com