How do I remove redundant elements from a Path?

By I Wayan Saryada in Core API, IO 0 Comment

To eliminate redundant elements from a Path we can use the Path.normalize() method. For example in the following code snippet. When try accessing the README file in the current directory the . symbol in the Path elements considered to be redundant, we don’t need it. That’s why we normalize the Path.

package org.kodejava.io;

import java.nio.file.Path;
import java.nio.file.Paths;

public class PathNormalize {
    public static void main(String[] args) {
        // The following Path contains a redundant element. The "." which 
        // basically point to the current directory can simply be removed
        // when we are working on the current directory.
        Path path = Paths.get("./README.md");
        System.out.println("Path = " + path);

        // Removes redundant name elements from the path.
        path = path.normalize();
        System.out.println("Path = " + path);
    }
}

How to get some information about Path object?

By I Wayan Saryada in Core API, IO 0 Comment

The java.nio.Path provides some methods to obtain information about the Path. For example, you can get information about the file name, the parent and the root path. For these you can call the getFileName(), getParent() and getRoot() method respectively.

You can also get the number of elements that make up this Path using the getNameCount() method. And to get the sub-path you can use the subpath() method and specify the starting and ending indexes. The code snippet below demonstrate to you how to get this information.

package org.kodejava.io;

import java.nio.file.Path;
import java.nio.file.Paths;

public class PathInfoExample {
    public static void main(String[] args) {
        // Create a Path for Windows notepad program.
        Path notepad = Paths.get("C:/Windows/System32/notepad.exe");

        // Get some information about the Path object.
        System.out.printf("File name         : %1$s%n", notepad.getFileName());
        System.out.printf("Name count        : %1$s%n", notepad.getNameCount());
        System.out.printf("Parent path       : %1$s%n", notepad.getParent());
        System.out.printf("Root path         : %1$s%n", notepad.getRoot());
        System.out.printf("Sub path from root: %1$s%n", notepad.subpath(0, 2));
    }
}

This code will print something like:

File name         : notepad.exe
Name count        : 3
Parent path       : C:\Windows\System32
Root path         : C:\
Sub path from root: Windows\System32

How do I create a java.nio.Path?

By I Wayan Saryada in Core API, IO 0 Comment

The following code snippet show you how to create a Path. A Path (java.nio.Path) in an interface that represent a location in a file system, such as C:/Windows/System32 or /usr/bin.

To create a Path we can use the java.nio.Paths.get(String first, String... more) methods. Below you can see how to create a Path by passing only the first string and by passing a first string plus some varargs string.

package org.kodejava.io;

import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;

public class PathCreate {
    public static void main(String[] args) {
        // Create a Path that represents Windows installation location.
        Path windows = Paths.get("C:/Windows");

        // Check to see if the path represent a directory.
        if (Files.isDirectory(windows)) {
            // do something
        }

        // Create a Path that represent Windows programs installation location.
        Path programFiles = Paths.get("C:/Program Files");
        Files.isDirectory(programFiles);

        // Create a Path that represent the notepad.exe program
        Path notepad = Paths.get("C:/Windows", "System32", "notepad.exe");

        // Check to see if the path represent an executable file.
        if (Files.isExecutable(notepad)) {
            // do something
        }
    }
}

How to use underscore in numeric literals?

By I Wayan Saryada in Basic, Core API 0 Comment

Writing a long sequence of numbers in a code is a hard stuff to read. In the new feature introduced by JDK 7 we are now allowed to write numeric literals using the underscore character to break the numbers to make it easier to read.

You can see how to use underscore in numeric literals in the following examples. And you’ll see it for yourself that it really makes numbers easier to read.

package org.kodejava.basic;

public class UnderscoreNumericExample {
    public static void main(String[] args) {
        // Write numeric literals using underscore as an easier way
        // to read long numbers.
        int maxInt = 2_147_483_647;
        int minInt = -2_147_483_648;

        if (maxInt == Integer.MAX_VALUE) {
            System.out.println("maxInt = " + maxInt);
        }

        if (minInt == Integer.MIN_VALUE) {
            System.out.println("minInt = " + minInt);
        }

        // Write numbers in binary or hex literals using the
        // underscores.
        int maxIntBinary = 0B111_1111_1111_1111_1111_1111_1111_1111;
        int maxIntHex = 0X7____F____F____F____F____F____F____F;

        System.out.println("maxIntBinary = " + maxIntBinary);
        System.out.println("maxIntHex    = " + maxIntHex);
    }
}

The results of the code snippet:

maxInt = 2147483647
minInt = -2147483648
maxIntBinary = 2147483647
maxIntHex    = 2147483647

How do I define an integer constant in binary format?

By I Wayan Saryada in Basic, Core API 0 Comment

The JDK 7 add a small feature to work with a binary number. In the previous JDK we have to use the Integer.parseInt() method if we need to work with other base number. But with this new feature introduced in the Project Coin we can simplify the code when we work with the binary number.

To specify a binary literal in the code, add the prefix 0b or 0B to the number. The following code snippet show you how to write the binary literals:

package org.kodejava.basic;

public class BinaryLiteralExample {
    public static void main(String[] args) {
        // In JDK 6 and the previous version you must use the
        // Integer.parseInt() method to define a number using
        // a binary literal.
        int x = Integer.parseInt("00101010", 2);
        System.out.println("x = " + x);

        // In the new JDK 7 you can simply use the following
        // binary literal to define a number using a binary
        // literal.
        int y = 0b00101010;
        System.out.println("y = " + y);
    }
}

The result of our code snippet:

x = 42
y = 42