Yes, that will produce two different values. This... x = (*(g)[*i]); ...is like this... x = (*((g)[*i])); Which is quite different from this... x = (*g)[*i]; HTH, --Eljay
Yes, that will produce two different values. This... x = (*(g)[*i]); ...is like this... x = (*((g)[*i])); Which is quite different from this... x = (*g)[*i]; HTH, --Eljay