What will be the output of the following C code?#include <stdio.h>int main(){int x = 2, y = 0;int z = (y++) ? y == 1 && x : 0;printf("%d\n", z);return 0;}012 Compile time error

The output of the given C code will be 0.

Here's the step by step explanation:

1. The variables x and y are initialized with the values 2 and 0 respectively.

2. The ternary operator is used to assign a value to z. The condition of the ternary operator is (y++), which checks if y is not zero. Since y is zero, the condition is false. Therefore, the value after the colon (which is 0) is assigned to z. The y++ operation increments y, but the value of y before the increment is used for the condition.

3. The printf function prints the value of z, which is 0.

4. The main function returns 0, indicating successful execution of the program.

So, the output of the program is 0.