how the answer become 15... any one can please explane in easy way. regards #include <iostream> using namespace std; int main() { int a = 2; int b= 1; cout<<(++a/b++)*5; | Sololearn: Learn to code for FREE!
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+ 1

how the answer become 15... any one can please explane in easy way. regards #include <iostream> using namespace std; int main() { int a = 2; int b= 1; cout<<(++a/b++)*5;

31st Oct 2016, 10:24 AM
zbwafi
4 Answers
+ 4
if we where to break this down it might make more sense. say we have this: int a = 2; cout << ++a<<endl; cout << a << endl; this will give us : 3 3 the reason for this is the ++ is used prefix which adds +1 to a. the increment happens BEFORE the variable is called. however, let's look at this: int b =1; cout <<b++<endl; cout << b <<endl; this will give us: 1 2 the reason for this is our first cout statement b is called BEFORE the increment. so b will be incremented however it will only be visible after that statement has finished. so putting it back together : cout << (++a/b++) * 5; is the same as cout << (a+1/b) * 5 for just that line
31st Oct 2016, 10:38 AM
Daniel Rollins
Daniel Rollins - avatar
+ 2
++a = 3, because the plus signs are in front of the variable. That means, that you first increment a by 1, before you start to calculate with it. b++ = 1, because the increment is done, after you use b in your calculation That means "cout<<(3/1)*5"
31st Oct 2016, 10:37 AM
Emre Ögretmen
Emre Ögretmen - avatar
+ 1
got it bro thank you.
31st Oct 2016, 10:39 AM
zbwafi
+ 1
thank you dear.
31st Oct 2016, 10:45 AM
zbwafi