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• Post category:Blog

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

```int var;
var = 2;
var = var * var;
var = var + var;
/*
var = var / var;
var = var % var;
*/```
• `8`
• `1`
• `0`
• `16`
 Explanation & Hints: To find the value of the variable `var` at the end of the given C code snippet, let’s walk through the code step by step: Initialization: ```int var; var = 2;``` Here, `var` is initialized to `2`. First Update: `var = var * var;` The value of `var` is squared, so `2 * 2 = 4`. Now `var` equals `4`. Second Update: `var = var + var;` The value of `var` is added to itself, so `4 + 4 = 8`. Now `var` equals `8`. The lines within the comment block are not executed, as they are commented out: ```/* var = var / var; var = var % var; */``` Thus, these lines do not affect the value of `var`. Therefore, the value of `var` at the end of this snippet is 8.

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

```int var;
var = 2;
var = var * var;
var = var + var;
var = var / var;
var = var % var;```
• `0`
• `1`
• `8`
• `16`
 Explanation & Hints: Let’s break down the operations step-by-step: ```int var; var = 2; var = var * var; var = var + var; var = var / var; var = var % var;``` Initial value: `var = 2` Multiplication: `var = var * var` `var = 2 * 2` `var = 4` Addition: `var = var + var` `var = 4 + 4` `var = 8` Division: `var = var / var` `var = 8 / 8` `var = 1` Modulo: `var = var % var` `var = 1 % 1` `var = 0` (since any number modulo itself is 0) So, the value of `var` at the end of the snippet is 0.

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