How Does This Code Print "Hello World" in Java?

What is the mechanism behind this Java code that prints "hello world"?

for (long l = 4946144450195624l; l > 0; l >>= 5) 
    System.out.print((char) (((l & 31 | 64) % 95) + 32));

The provided Java code snippet prints the string "hello world" by manipulating a long integer using bitwise operations and character encoding. Let's break down the code to understand how it achieves this outcome.

for (long l = 4946144450195624L; l > 0; l >>= 5) {
    System.out.print((char) (((l & 31 | 64) % 95) + 32));
}

Causes

• The initial value of the long variable `l` is 4946144450195624L, which in binary has specific bits set that will ultimately map to ASCII characters.
• The loop runs while `l` is greater than zero, right-shifting the bits of `l` by 5 on each iteration. This means that each iteration processes a different set of bits from `l`.
• The expression inside the `print` method converts bits from `l` into valid ASCII character codes by manipulating them with bitwise operations.

Solutions

• The operation `(l & 31 | 64)` isolates specific bits of `l`, using bitwise AND to mask bits and then using bitwise OR to set the 7th bit (which corresponds to uppercase letters in ASCII).
• The modulo operation `% 95` ensures the result remains within the ASCII printable character range before adding 32, which shifts the code to the printable ASCII character range (from 32 to 126).
• The final cast to `(char)` converts the resulting integer into its corresponding ASCII character, and this character is printed to the console.

Mistake: Misunderstanding bitwise operations in Java, leading to incorrect assumptions about character output.

Solution: Review the basic principles of bitwise operations and how they manipulate integers.

Mistake: Failing to recognize that right shifting (`>>=`) affects the number of bits processed in each iteration.

Solution: Observe how bitwise right shift divides the number by powers of two, effectively narrowing down the bits processed.