Printable View
When i run the following code
main()
{
int i=-3,j=2,k=0,m;
m=++i&&++j||++k;
printf("\n %d %d %d %d " ,i,j,k,m);
}
the result i get is
-2 3 0 1
What i'm not able to understand is that why the value of k is not incremented??
I was trying different things with the expression , i noticed that if i use m=++i&&++j&&++k; instead of the original expression the result is
-2 3 1 1
i.e. this time k has been incremented ...WHY??
My best gues is that the system is doing early termination of the conditionals. Once the ++i&&++j resolve to true, it doesn't matter what k is because the result and 1 || (1 or 0) is always 1. So the check is terminated and k is never incremented. When you replace || with &&, the system must evaluate k to determine the results.
PK
> My best gues is that the system is doing early termination of the conditionals.
You are right, but you shouldn't be guessing such things.
> m=++i&&++j||++k;
Which is why in most programming standards (and professional code), mixing operators with side effects with other operators is avoided.
Sure it's legal code, but if you have to sit and think about it for several minutes each time you see it, it's hardly productive code.
Well yeah, but even brain dead monkey who just learned how to program wold think thatwold be replaceble withCode:m = ++i && ++j || ++k;
But it isn't! :mad: I blame god!Code:i++;
j++;
k++;
m = i && j || k;
Doesn't learning how to program involve reading? Every C book I've ever read talks about early termination with conditional operators (if the book talks about conditional operators).Quote:
Originally Posted by iwabee
Ok monkey was bad example, but I wasn't talking about early termination of whatever. Rather of the whole concept of increment/decrement operators. Increment before/after use VS. increment before/after statement.
Thanx to all........