Java /检查异常中的异常处理

Question

I am new to Java, and would appreciate any help in the following program, where I am trying to provide for a checked exception .

Her's example:

package book1;
import java.io.*;

public class CheckedException {

public static void main(String[] args) {
openFile("D:Java.txt");

} 

public static void openFile(String name)
{
try
{
FileInputStream f=new FileInputStream(name );
}
catch(FileNotFoundException e)
{
System.out.println("File Not Found!!");
}
}
}

The problem:

1. The ideal path to a text file on Windows should be D:\\file.txt. But in this program, this syntax gives an error, indicating an incorrect escape syntax(\\n,\\t etc). Why is the compiler(I'm using Eclipse Kepler on Windows 8.1) treating the path file as an escape seq?

2. Even when I removed the \\ from the notation(path D:java.txt), the program throws the file not found exception, and is subsequently caught which displays the file not found message.

Please help,

thank you.

Answer 1

In java (and many other languages), you must escape the back slash with another backslash when coding String literals:

openFile("D:\\Java.txt"); // This is the string D:\Java.txt

but java understands forward slashes in file paths on operating systems that use backslashes for separators, so you could just do this:

openFile("D:/Java.txt");

---

To explain the error you're seeing, the string literal "\\:" is an "escaped colon", which is not a valid java string escape sequence.