Given: 11. String test = "This is a test"; 12. String[] tokens = test.split("\s"); 13. System.out.println(tokens.length); What is the result?
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0
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1
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4
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Compilation fails
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An Exception thrown at runtime
If the string literal is "\s" (single backslash), this is invalid because \s is not a recognized escape sequence in Java. String escape sequences like \n, \t, \" must be valid. To represent a regex backslash, you need "\\s" (double backslash in source = single backslash in string).
To answer this question, let's go through each line of code and understand what it does:
Line 11: String test = "This is a test"; - This line declares a string variable named "test" and assigns it the value "This is a test".
Line 12: String[] tokens = test.split("\s"); - This line splits the string "test" into an array of substrings based on the delimiter specified as "\s". However, there is an issue with the delimiter used. "\s" is not a valid regular expression for whitespace in Java. To represent a whitespace character, you should use "\s" instead. Therefore, this line would result in a compilation error.
Line 13: System.out.println(tokens.length); - This line prints the length of the "tokens" array. However, since there was a compilation error in line 12, the code would not compile successfully.
Therefore, the correct answer is D) Compilation fails.