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Q[StringBuffer]
The code snippet given below will result in:
StringBuffer x = new StringBuffer("Hello");
StringBuffer y = new StringBuffer("HELLO");
System.out.println(x.equalsIgnoreCase(y));

ID: #24936 Competency focused Practice Questions ISC Class XII Computer Science 5 views

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#24936Q ID

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Competency focused Practice Questions ISC Class XII Computer ScienceTopic

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A boolean value false ✔ ✖
B boolean value true ✔ ✖
C undeclared method equalsIgnoreCase(...) ✔ ✖
D String literals mismatch ✔ ✖

Correct Answer: Option C

Explanation

StringBuffer MCQ Question

StringBuffer x = new StringBuffer("Hello");
StringBuffer y = new StringBuffer("HELLO");
System.out.println(x.equalsIgnoreCase(y));

Correct Answer: (c) undeclared method equalsIgnoreCase(...)

Explanation:

In Java, the class StringBuffer does NOT have a method called equalsIgnoreCase().

The method equalsIgnoreCase() belongs to the String class, not StringBuffer. Therefore, calling this method on a StringBuffer object will cause a compile-time error.

Java checks method availability at compile time. Since x and y are StringBuffer objects, the compiler cannot find equalsIgnoreCase(), resulting in an error.

So the program does not run at all, and the correct answer is:
undeclared method equalsIgnoreCase(...)

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Q[StringBuffer] The code snippet given below will result in: StringBuffer x = new StringBuffer("Hello"); StringBuffer y = new StringBuffer("HELLO"); System.out.println(x.equalsIgnoreCase(y));