Java ProgramsControl FlowCheck Vowel Using Switch

Check Vowel Using Switch in Java

beginner·  Control Flow  ·  Switch Statement

Problem

Stacking every vowel — both cases — under one switch block lets a single assignment cover all ten matching characters at once.

Given a single character, determine whether it is a vowel or a consonant.

Input
ch = 'o'
Output
'o' is a vowel

Java Program

Java
public class VowelCheckSwitch { public static void main(String[] args) { char ch = 'o'; boolean isVowel; switch (ch) { case 'a': case 'e': case 'i': case 'o': case 'u': // lowercase vowels case 'A': case 'E': case 'I': case 'O': case 'U': // uppercase vowels isVowel = true; break; default: isVowel = false; } System.out.println("'" + ch + "' is a " + (isVowel ? "vowel" : "consonant")); } }

Output

'o' is a vowel

Core Logic

Stacking all ten vowel characters — both cases — under a single case block, with every non-vowel falling through to default, decides vowel or consonant in one switch.

How It Works
  1. 1case 'a': case 'e': ... case 'U': stacks all five vowels in both lowercase and uppercase under one block.
  2. 2Any character matching one of those ten labels falls into the shared block and sets isVowel = true.
  3. 3The default case catches everything else and sets isVowel = false.
For ch = 'o', it matches one of the stacked vowel labels, so isVowel becomes true.
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Key Point: Stacking ten case labels with no code between them is what lets a single isVowel = true; cover every vowel — the labels share the block precisely because they share the same outcome.

Key Concepts

switch statementstacked case labelsdefault case

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