Programming


Java: Breakdown an integer to the powers of two it is composed from 1

The following function will accept an integer number as input and will produce an IntStream, which is a sequence of primitive int-valued elements that supports sequential and parallel aggregate operations.

public static IntStream splitToPowersOfTwo(final int input) {

    int temp = input;
    final IntStream.Builder builder = IntStream.builder();

    while (temp != 0) {

        //Integer.lowestOneBit: Finds the lowest-order ("rightmost") one-bit in the specified int value and returns an int number that has only the bit in the previously matched position set as 1. This value is a power of two that is one of the components of the number. If the number is zero, the function returns zero.
        final int powerOfTwo = Integer.lowestOneBit(temp);
        builder.add(powerOfTwo);
        //We update our temp by subtracting the number we got, our goal is to remove the previously matched bit and get the next one on the next iteration.
        temp = temp & ~ powerOfTwo;
    }

    return builder.build();
}

We used Integer.lowestOneBit as it makes it easy for us to breakdown the number to its power of two components by matching for us only one bit at time.

Example of usage:

Scenario: We want to breakdown an integer to the powers of two that their sum produces the number and create a String with those values. Using our function, we can do the following:

final String output = splitToPowersOfTwo(flags).mapToObj(Integer::toString).collect(Collectors.joining(", "));

Note: an IntStream is not the same as a Stream<Integer>, we used .mapToObj(Integer::toString) which calls the static method Integer.toString(int). This is different to calling .map(Integer::toString) on a Stream<Integer> as the latter won’t compile because it is ambiguous.


Java: Given a path to a file, create the folders to it

Assuming you have a full path to a file you want to create but the folder to it might exist or not, you can use the following code in your program to check if the folder exists and if it doesn’t, create it.

//The file we want to create;
final String logFileName = "./someFolder/anotherFolder/etc/myFile.txt";
//We find the last occurrence of the File.separator value, which will allow us to separate the path from the filename
final int lastIndex = logFileName.lastIndexOf(File.separator);
//The actual path we want to make sure it exists
final String dirPath = logFileName.substring(0, lastIndex > 0 ? lastIndex : logFileName.length());
//We try to use the folder by assigning it to a File object
final File theDir = new File(dirPath);
//Make the check that the folder exists
if (!theDir.exists()) {
    //If the folder does not exist, create it
    theDir.mkdirs();
}
//Continue with execution

C++: Get size of enum 1

We used the following application in C++ to test the size of an enum:

#include <iostream>
using namespace std;

enum SINGLE {
        S_ZERO = 0x0,
        S_FULL = 0xFFFFFFFF
};

enum DOUBLE {
        D_ZERO = 0x0,
        D_FULL = 0xFFFFFFFFFFFFFFFF
};

int main() {
        cout << "Single Zero: Size '" << sizeof(S_ZERO) << "' Value '" << S_ZERO << "'" << endl;
        cout << "Double Zero: Size '" << sizeof(D_ZERO) << "' Value '" << D_ZERO << "'" << endl;

        cout << "Single Full: Size '" << sizeof(S_FULL) << "' Value '" << S_FULL << "'" << endl;
        cout << "Double Full: Size '" << sizeof(D_FULL) << "' Value '" << D_FULL << "'" << endl;
        return 0;
}

The output we got is the following:

Single Zero: Size '4' Value '0'
Double Zero: Size '8' Value '0'
Single Full: Size '4' Value '4294967295'
Double Full: Size '8' Value '18446744073709551615'

From the result it is pretty easy to understand that the size of an enum will grow to 64bit when any of its values is greater than 32bit.

For our test we used g++ (GCC) 5.3.1 20160406 (Red Hat 5.3.1-6) on a 64bit Fedora 23.


PHP: Convert JavaScript-escaped Unicode characters to HTML hex references

There are cases where one might receive in PHP, escaped Unicode characters from the client side JavaScript. According to the RFC it is normal for JavaScript to convert characters to that format and in effect that we receive any character in the escaped format of \uXXXX in PHP.

Any character may be escaped.
If the character is in the Basic Multilingual Plane (U+0000 through U+FFFF),
then it may be represented as a six-character sequence:
a reverse solidus, followed by the lowercase letter u, followed by four hexadecimal digits that encode the character's code point.
The hexadecimal letters A though F can be upper or lowercase.

A sample input you might receive could look like this George\u2019s treasure box instead of George’s treasure box.

This kind of input should not be stored as is as it does not make sense to the HTML language, instead we should fix it up using preg_replace.

$decoded = preg_replace('/\\\\u([a-fA-F0-9]{4})/', '&#x\\1;', $input);

The above command will look for all instances of \uXXXX in the $input and it will replace each one with the appropriate character using the XXXX value that it will match.

What this part '/\\\\u([a-fA-F0-9]{4})/' of the code do is the following:

  • \\\\ – Find the character \ in the string, the reason we have four \ instead of one, is because it has special meaning in the regular expression and we have to escape it. For that reason we need to use two of them and get \\. After that, we need to escape each of them again due to the special meaning they have in PHP and we end up with four of them.
  • u – The previous step must be followed by a u character.
  • ([a-fA-F0-9]{4}) – After the previous step has matched, we need to match 4 characters. Each of them must be either a character from A-Z or a-z or 0-9.

This part '&#x\\1;' will:

  • &#x – Is a constant string that will print the characters &#x. These characters will instruct HTML to print the character that will occur using hexadecimal entity reference that will follow.
  • \\1 – Contains the reference of the 1st parenthesized pattern. In this case we only have a parenthesis around the XXXX part of the \uXXXX so \\1 will be replaced with the XXXX value.