Confusing increment operator

Question

int a=10;
a = a++ + 10;
printf("%d",a);
Output
Is 20
Please explain how

Answer 1

Siddharth Jain , it is because the post increment "++" operator. Post increment first evaluate the expression then increment it. It's exactly the opposite with pre increment => if you put "++" before the variable - first increment the value and then evaluate the expression. You can change it and you'll see the difference => with pre increment it's 21, post increment it is 20. Hope you understand it better 😉

Answer 2

TheWh¡teCat can u explain this

Answer 3

TheWh¡teCat but what about the increment operator after assignment

Answer 4

It's because the value was assigned to "a" before the increment happened.

Answer 5

a++ is postfix.
The postfix form uses the value of the variable first, before incrementing/decrementing it.
If you change the code a bit, then an example of incriment postfix a ++ will be more obvious:
int a = 10;
int b = 0;
b = a++ + 10; /* step 1: b=10+10;
step 2: a=10+1; */
printf("%d",a); //Output Is 11
printf("%d",b); //Output Is 20
https://www.sololearn.com/learn/C/2917/