What will be the output of the following code ?#include<stdio.h>int main(){int x=10,y;y=x++ + ++x;printf("%d %d %d %d ",y,x++,x,++x);}

The output of the code will be "32 13 14 15".

Here's the step-by-step explanation:

1. The initial value of x is 10.
2. In the expression y=x++ + ++x;, the post-increment operator (x++) is used first, which means the value of x is used in the expression before it is incremented. So, for this part, x is still 10. After this operation, x is incremented by 1, making x = 11.
3. The pre-increment operator (++x) is used next, which means the value of x is incremented before it is used in the expression. So, x is incremented by 1 again, making x = 12, and then used in the expression.
4. Therefore, y = 10 (from the first x++) + 12 (from the second ++x) = 22.
5. The printf statement prints the value of y, which is 22.
6. Then, it prints the value of x after it is incremented by the post-increment operator (x++), which makes x = 13.
7. Next, it prints the current value of x, which is 13.
8. Finally, it prints the value of x after it is incremented by the pre-increment operator (++x), which makes x = 14.