+ 3
% <--
This symbol calculates to remainder ONLY of two #'s right? ex.) cout<<9%3; //output= 0 ! 3 cout<<7%2; //output= 5 ! 3 or am I doing this all wrong?
2 ответов
+ 6
Yes remainder only.
9%2=1
7%4=3
1001%11=0
+ 5
Well, almost. It basically just takes the remainder of the division of the numbers, but only for the order that they are set.
For example, the equation 8%3 will only return 2 because 8/3 equals 2 which leaves a remainder of 2 (2•3=6, 8-6=2). It will not do this the other way around, so in the example it will not return 5.