Evaluate the following expressions and get the value of a, b, c and d by hand.x = 10;y = 20;z = 30;a = x++ + 10;b = --x – x--;c = ++x - ++y - ++z;d = ++y * ++y;Write a C program

Let's evaluate the expressions one by one:

1. a = x++ + 10; Here, x++ is a post-increment operation which means it will increment the value of x by 1 after the operation. So, x becomes 11 after this operation but the value of x used in the operation is 10. Hence, a = 10 + 10 = 20.

2. b = --x – x--; Here, --x is a pre-decrement operation which means it will decrement the value of x by 1 before the operation. So, x becomes 10 before the operation and hence b = 10 - x--. Now, x-- is a post-decrement operation which means it will decrement the value of x by 1 after the operation. So, x becomes 9 after this operation but the value of x used in the operation is 10. Hence, b = 10 - 10 = 0.

3. c = ++x - ++y - ++z; Here, ++x, ++y and ++z are pre-increment operations which means they will increment the value of x, y and z by 1 before the operation. So, x becomes 10, y becomes 21 and z becomes 31 before the operation. Hence, c = 10 - 21 - 31 = -42.

4. d = ++y * ++y; Here, ++y is a pre-increment operation which means it will increment the value of y by 1 before the operation. So, y becomes 22 before the operation and hence d = 22 * 22 = 484.

Now, let's write a C program to verify these results:

#include <stdio.h>

int main() {
    int x = 10;
    int y = 20;
    int z = 30;
    int a, b, c, d;

    a = x++ + 10;
    b = --x - x--;
    c = ++x - ++y - ++z;
    d = ++y * ++y;

    printf("a = %d\n", a);
    printf("b = %d\n", b);
    printf("c = %d\n", c);
    printf("d = %d\n", d);

    return 0;
}

When you run this program, it will print:

a = 20
b = 0
c = -42
d = 484

which verifies our manual calculations.