How the output of this preincrement and postincrement equation in c is 2,2?

Question

#include <stdio.h>

int main(){
int i = 0;
int j = 0;
j = i++ + ++i;
printf("%d %d", i,j);
}
output = 2, 2

here ++i = 1
then 1+i(0) = 1,
hence j should be = 1, so why j = 2?

and if i++ + ++i = 1+1 = 2 = j
then, why i = 2?

someone help me with this please.

Answer 1

Radib 77 this seems a question on pre increment or post increment understanding

Unfortunately, this problem is not well drafted. Generally in real world , you will not come across such situation where in same line you do pre and post increment on a single variable....

Still doing so , it results in undefined behaviour. Code behavior depends on compiler to compiler and even same compiler gives different answer at different time.

Here on your expected output case , it is evaluated like below :

++i is resulted to i as 1 as ++ has highest priority

Then i+i is 1+1 which means 2 is assigned to j

Once this line is executed, i ++ is evaluated making i from 1 to 2

Answer 2

Ketan Lalcheta thanks man. Understood thank you.

Answer 3

Rather than speculating on how and why, search the net for more information around 'sequence point' and 'expression evaluation' 👍

Answer 4

preincrement (++i) increment variable before evaluating, while postincrement (i++) increment after...

j = i++ + ++i

is equal to 0 (post increment) + 2 (pre increment), because the second access to i is done after first post increment (1) and before evaluating (2)