C#.NET — Evaluate bitwise OR of character literals and switch grouping result. char ch = Convert.ToChar('a' | 'b' | 'c'); switch (ch) { case 'A': case 'a': Console.WriteLine("case A | case a"); break; case 'B': case 'b': Console.WriteLine("case B | case b"); break; case 'C': case 'c': case 'D': case 'd': Console.WriteLine("case D | case d"); break; }

Introduction / Context:
This tests understanding of bitwise OR on character codes and how a switch with grouped cases routes execution to a specific write line.

Given Data / Assumptions:

• Character literals: 'a' = 97, 'b' = 98, 'c' = 99 (ASCII/Unicode code points in this range).
• Expression: Convert.ToChar('a' | 'b' | 'c').
• Switch groups map to different Console.WriteLine strings.

Concept / Approach:
Bitwise OR combines set bits. Since 97|98 == 99, and 99|99 == 99, the result is 'c'. The switch has a group that includes 'C'/'c' and also routes to a print line labeled "case D | case d".

Step-by-Step Solution:

Verification / Alternative check:
Write hex: 0x61 | 0x62 = 0x63; 0x63 | 0x63 = 0x63; 0x63 corresponds to 'c'.

Why Other Options Are Wrong:

• "case A | case a" and "case B | case b": not selected for 'c'.
• Compile Error: valid code.
• No output: one of the cases matches; printing occurs.

Common Pitfalls:
Assuming grouped cases must print their own letter; in this example, the author intentionally prints the string mentioning D for multiple labels.

Final Answer:
case D | case d