#include <iostream> using namespace std; int main() { int a=10,b; b=++a+(++a); cout<<b; return 0; }

Question

why I am getting 24 as a output please help thanks..

Answer 1

Yesterday, i posted same question like this.
int a=5,b=0;
b=++a + ++a + ++a;
cout<<b;

The output is 22. Here's my explanation.

First of all we divide the expression in 3 parts ++a each. When first ++a is performed the value of "a" becomes 6, but it's not stored into stack, when second parts is performed the values becomes 7.This time addition of both is performed, but as the value of a is 7, the values stored into stack are 7,7. Now the last part is evaluated & the value of "a" becomes 8 & as there is addition to be performed, it is stored into stack. So the contents of stack are 7,7,8. By performing addition of them answer comes out to be 7+7+8=22.
When we use "volatile int", each time when value of "a" increments, it's stored into the stack. So the stack content in this case is 6,7,8 & answer comes out to be 6+7+8=21.
Note that the logic is followed in GCC platform. If you're using Turbo C, then you will get the answer as 24. In this case stack contents are same as last value i.e. 8,8,8.🙂

Answer 2

If you want to have 22 :
https://code.sololearn.com/c4C8jY3KVvRU/?ref=app

Answer 3

No problem bro✌

Answer 4

because of the use of non volatile int.
change the line to-

volatile int a=10, b;

to get the desired output.

Answer 5

can we use volatile keyword in C language also..

Answer 6

thanks

Answer 7

For a better understanding see this code-

https://code.sololearn.com/ctog8Gtjyt0f/?ref=app

Answer 8

Yes

Answer 9

thanks bro it really helped I was confused..