使用Java 8，打印文件中所有行的最优选和简洁方法是什么？

Question

What is the most preferred and concise way to print all the lines in a file using the new Java 8?

The output must be a copy of the file, line for line as in:

[Line 1 ...]
[Line 2 ...]

The question pertains to Java-8 with lambdas even though there are ways of doing it in older versions. Show what Java-8 is all about!

Answer 1

This uses the new Stream with a lambda in a try enclosure.

I would say it is the most preferred and concise way because:

1) It will automatically close the stream when done and properly throw any exceptions.

2) The output of this is lazy. Each line is read after the last line is processed. This is also is closer to the original Java streams based file handling spec.

3) It prints each line in a manner that most closely resembles the data in the file.

4) This is less memory intensive because it does not create an intermediate List or Array such as the Files.readAllLines(...)

5) This is the most flexible, since the Stream object provided has many other uses and functions for working with the data (transforms, collections, predicates, etc.)

 try (Stream<String> stream = Files.lines(Paths.get("sample.txt"),Charset.defaultCharset())) {
            stream.forEach(System.out::println);
 }

If the path and charset are provided and the Consumer can take any Object then this works too:

 try (Stream stream = Files.lines(path,charset)) {
            stream.forEach(System.out::println);
 }

With error handling:

 try (Stream<String> stream = Files.lines(Paths.get("sample.txt"),Charset.defaultCharset())) {
            stream.forEach(System.out::println);
 } catch (IOException ex) {
        // do something with exception
 } 

Answer 2

You don't really need Java 8:

System.out.println(Files.readAllLines(path, charset));

If you really want to use a stream or need a cleaner presentation:

Files.readAllLines(path, charset).forEach(System.out::println);

Answer 3

这是一个简单的解决方案：

System.out.println( new String(Files.readAllBytes(Paths.get("sample.java"))) );