What is the value of the k variable at the end of the following snippet?

To determine the value of the variable k at the end of the provided C code snippet, we will evaluate each of the expressions in the assignment to k based on the values of i and j. Here’s the detailed evaluation:

Variable Initialization:

int i, j, k;
i = 3;
j = -3;

k = (i >= i) + (j <= j) + (i == j) + (i > j);

k = 1 + 1 + 0 + 1;
k = 3;

Thus, the value of k at the end of this code snippet is 3.

Let’s evaluate the code snippet step by step to find the value of the variable k:

Initialization of Variables:

int i, j, k;
i = 3;
j = -3;

k = i * j;

k += j;

k /= i;

Let’s break down the operations step-by-step:

int i, j, k;

i = 4;
j = 5;

k = i-- * ++j;

1. Initial values:
   • i = 4
   • j = 5
2. Pre-increment j (++j):
   • ++j increments j by 1 before using its value.
   • So, j becomes 6.
3. Post-decrement i (i--):
   • i-- uses the current value of i (which is 4) and then decrements i by 1.
   • So, the value of i used in the multiplication is 4, and after the operation, i becomes 3.
4. Multiplication:
   • k = i-- * ++j
   • k = 4 * 6
   • k = 24

So, the value of k at the end of the snippet is 24.

Let’s break down the operations step-by-step:

int i, j, k;

i = 4;
j = 5;

k = --i * j++;

1. Initial values:
   • i = 4
   • j = 5
2. Pre-decrement i (--i):
   • --i decrements i by 1 before using its value.
   • So, i becomes 3.
3. Post-increment j (j++):
   • j++ uses the current value of j (which is 5) and then increments j by 1.
   • So, the value of j used in the multiplication is 5, and after the operation, j becomes 6.
4. Multiplication:
   • k = --i * j++
   • k = 3 * 5
   • k = 15

So, the value of k at the end of the snippet is 15.