Java Program to Print a Semicolon Without Using a Semicolon

In this Java program, we will learn how to print a semicolon (;) without using a semicolon anywhere in the program.

This is a common Java programming puzzle and interview question that demonstrates how Java expressions can be used inside control statements.

Java Program

public class PrintSemicolon {
    public static void main(String[] args) {
        if (System.out.printf("%c", 59) != null) {
        }
    }
}

Output

;

Explanation

The ASCII value of the semicolon character (;) is 59.

;  →  ASCII value = 59

We use the following statement:

System.out.printf("%c", 59)

The %c format specifier tells printf() to print the value as a character. Therefore, 59 is printed as the semicolon character.

The interesting part is that we don’t write the printf() call as a normal statement. Instead, we place it inside an if condition:

if (System.out.printf("%c", 59) != null) {
}

An if condition is an expression and does not require a semicolon at the end. printf() returns a PrintStream object, so its result can be compared with null.

As a result, the semicolon is printed without using a semicolon to terminate the statement.

Key Points

• The ASCII value of ; is 59.
• %c is used to print a character.
• System.out.printf() returns a PrintStream object.
• The printf() call is placed inside an if condition.
• The program contains no semicolon character (;) in the source code.

Note: This is a programming puzzle rather than a recommended coding practice. In normal Java programs, semicolons should be used wherever the Java syntax requires them.