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

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### Basic data types, operations, and flow control (decision-making statements) Module 2 Test Answers Full 100%

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

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### Basic data types, operations, and flow control (decision-making statements) Module 2 Test Answers Full 100%

## 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 |

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### Basic data types, operations, and flow control (decision-making statements) Module 2 Test Answers Full 100%

## 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 |