How does this code work(C)?

Question

I got this in challenge today but did not understand how it works?

int x = 2, y = 0;
int z = (y++) ? 2 : y == 1 && x;
printf("%d", z);
//Output:1

Answer 1

First y is initially 0 so due to post increment which has hugest precedence in expression due to parenthesis of ternary so it evaluate first, post increment store value of y first which is 0 is replaced in the parenthesis( y++) place and when y again comes to use it evaluate the increment and value of y will become 1
So expression after first step is
int z = 0 ? 2 : 1 == 1 && 2;

Now == has highest priority in precedence so it evaluated and 1== 1 so it will return true

int z = 0 ? 2 : true && 2;

Now && operator has next priority in precedence so && will be evaluated so two non zero and true value return true so return value by && is 1

int z = 0 ? 2 : 1

Now as ternary left to evaluate so it is evaluated as 0 is false so second expression is choose in ternary which is 1 so final output is returned as 1

Answer 2

y++ is 0 so y==1 && x is evaluated.
y is 1 now and y==1 is true and x is a positive value so the expression returns 1 since it is both true.
Thus the output 1.

Answer 3

y++ means use the value first and then increment so it will be 0 when you use it first and will become 1 when you use it 2nd time.

Answer 4

Avinesh Thanks for the explaination👍
GAWEN STEASY Thanks, but your explaination is a little confusing.

Answer 5

Mohammed Niems Go to menu and click copy text.