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

int i, j, k;

i = 3;
j = -3;

k = (i >= i) + (j <= j) + (i == j) + (i > j);
• 3
• 2
• 1
• 0
 Explanation & Hints: 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; Assignment to k: k = (i >= i) + (j <= j) + (i == j) + (i > j); Breaking down each part: (i >= i): This checks if i is greater than or equal to i. Since i is equal to itself, this is true, and thus, evaluates to 1. (j <= j): This checks if j is less than or equal to j. Since j is equal to itself, this is true, and thus, evaluates to 1. (i == j): This checks if i is equal to j. Since 3 is not equal to -3, this is false, and thus, evaluates to 0. (i > j): This checks if i is greater than j. Since 3 is greater than -3, this is true, and thus, evaluates to 1. Summing these results gives: k = 1 + 1 + 0 + 1; k = 3; Thus, the value of k at the end of this code snippet is 3.

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

int i, j, k;

i = 3;
j = -3;

k = i * j;
k += j;
k /= i;
• -4
• -8
• 4
• 8
 Explanation & Hints: 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; Calculation of k: First, k is calculated by multiplying i and j: k = i * j; Substituting the values, =3×−3=−9. Next, j is added to k: k += j; This translates to =−9+(−3)=−12. Finally, k is divided by i: k /= i; This results in =−12÷3=−4. Therefore, the value of k at the end of this snippet is -4.

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

int i, j, k;

i = 4;
j = 5;

k = i-- * ++j;
• 24
• 28
• 21
• 18
 Explanation & Hints: Let’s break down the operations step-by-step: int i, j, k; i = 4; j = 5; k = i-- * ++j; Initial values: i = 4 j = 5 Pre-increment j (++j): ++j increments j by 1 before using its value. So, j becomes 6. 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. Multiplication: k = i-- * ++j k = 4 * 6 k = 24 So, the value of k at the end of the snippet is 24.

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

int i, j, k;
i = 4;
j = 5;

k = --i * j++;
• 15
• 18
• 16
• 12
 Explanation & Hints: Let’s break down the operations step-by-step: int i, j, k; i = 4; j = 5; k = --i * j++; Initial values: i = 4 j = 5 Pre-decrement i (--i): --i decrements i by 1 before using its value. So, i becomes 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. Multiplication: k = --i * j++ k = 3 * 5 k = 15 So, the value of k at the end of the snippet is 15.

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