C#.NET — Which snippets correctly classify an integer as Odd or Even (consider operators and definite assignment)?

Difficulty: Easy

Correct Answer: None of these

Explanation:


Introduction / Context:
The task is to identify which code fragments are valid C# for printing or storing whether an integer a is "Odd" or "Even". Beyond logic, you must check language details such as supported operators and definite assignment rules for local variables.


Given Data / Assumptions:

  • Snippets 1 and 4 declare int a; without assigning a value before using it.
  • Snippets 2 and 3 use the Mod operator keyword.
  • We evaluate both syntax and compile-time rules.


Concept / Approach:
In C#, a local variable must be definitely assigned before its value is read. Also, C# uses the % operator for remainder; the VB-style Mod keyword is not a C# operator. Therefore, even if the logic looks fine, using an uninitialized local or an unsupported operator makes a snippet invalid.


Step-by-Step Solution:

Snippet 1: uses a % 2 before a is assigned → compile-time error (definite assignment). Snippet 2: uses a Mod 2 → invalid operator in C#. Snippet 3: Console.WriteLine(a Mod 2 == 0 ? "Even" : "Odd") → again uses Mod → invalid. Snippet 4: conditional expression assigns to res in both branches, but still reads a before a is assigned → compile-time error.


Verification / Alternative check:
Fixing any snippet requires (a) assigning a first, e.g., int a = 7;, and (b) replacing Mod with % where present.


Why Other Options Are Wrong:

  • A/B/C/D claim one or more invalid snippets are correct; each relies on uninitialized locals or the non-existent Mod operator in C#.


Common Pitfalls:
Assuming VB syntax applies to C#, or overlooking the compiler’s definite assignment checks for locals.


Final Answer:
None of these

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