[OCPJP]Path和Files的用法

Given the code fragment:

Path file = Paths.get("courses.txt");
// line n1

Assume the courses.txt is accessible.

Which code fragment can be inserted at line n1 to enable the code to print the content of the courses.txt file?

A.

List<String> fc = Files.list(file);
fc.stream().forEach (s -> System.out.println(s));

B.

Stream<String> fc = Files.readAllLines(file);
fc.forEach (s -> System.out.println(s));

C.

List<String> fc = readAllLines(file);
fc.stream().forEach (s -> System.out.println(s));

D.

Stream<String> fc = Files.lines(file);
fc.forEach (s -> System.out.println(s));

題解

選項A，Files類別的list方法會回傳Stream物件，所以會編譯錯誤。

選項B，Files類別的readAllLines方法會回傳List物件，所以會編譯錯誤。

選項C，readAllLines方法前應該指定Files類別。

選項D，正確答案。