Given the code fragment:
What is the result?
Exam 1z0-809 All QuestionsBrowse all questions from this exam
Question 101
Correct Answer: D
The code results in a compilation error at line n2 because the UnaryOperator<String> requires a String input, but strBuf is of type StringBuffer. The method apply for UnaryOperator<String> expects a String argument, and passing a StringBuffer causes a type mismatch.
Discussion
jduarteOption: D
Answer is D. The method apply(String) in the type Function<String,String> is not applicable for the arguments (StringBuffer). c.appy(strBuff) is not correct.
HanenBAOption: D
UnaryOperator<String> c accept a string as input not StringBuffer, so c.apply(strBuf) will cause a compilation error
Tarik2190Option: D
Answer is D, public class Test { public static void main (String[] args) { final String str1 = "Java"; StringBuffer strBuf = new StringBuffer("Course"); UnaryOperator<String> u = (str2) -> str1.concat(str2); UnaryOperator<String> c = (str3) -> str3.toLowerCase(); System.out.println(u.apply(strBuf)); } }
pul26Option: D
Answer is D strBuf is not a String
Svetleto13Option: D
D,tested
asdfjhfgjuaDCVOption: D
D is the correct answer
steefaandOption: D
Answer is D. UnaryOperator takes a String and not StringBuffer.
hmcbqOption: A
Answer is A. If lambda has parenthesis than parameter should be with type.