Type promotion in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of arithmetic with byte and int

What is the output of this Java code snippet?

Java

public class Main {
    public static void main(String[] args) {
        byte b = 10;
        int i = 20;
        var result = b + i;
        System.out.println(result);
    }
}

A 30.0

B 30

C Compilation error

D Runtime error

💡 Hint

Remember that byte is promoted to int before arithmetic.

In Java, when a byte is used in arithmetic with an int, the byte is promoted to int. So b + i is int + int, resulting in int 30.

❓ Predict Output

intermediate

Output of char and int addition

What is the output of this Java code snippet?

Java

public class Main {
    public static void main(String[] args) {
        char c = 'A';
        int i = 1;
        int result = c + i;
        System.out.println(result);
    }
}

A 66

B B

C Compilation error

D Runtime error

💡 Hint

char is promoted to int before addition.

The char 'A' has Unicode value 65. Adding 1 results in 66, which is printed as an int.

❓ Predict Output

advanced

Output of compound assignment with byte

What is the output of this Java code snippet?

Java

public class Main {
    public static void main(String[] args) {
        byte b = 5;
        b += 10;
        System.out.println(b);
    }
}

A 5

B Compilation error

C Runtime error

D 15

💡 Hint

Compound assignment operators include implicit casting.

The compound assignment operator '+=' automatically casts the result back to byte, so no error occurs and b becomes 15.

❓ Predict Output

advanced

Output of simple assignment with byte and int addition

What is the output or error of this Java code snippet?

Java

public class Main {
    public static void main(String[] args) {
        byte b = 5;
        b = b + 10;
        System.out.println(b);
    }
}

A Compilation error

B 15

C Runtime error

D 5

💡 Hint

Check the type of b + 10 expression and assignment compatibility.

b + 10 promotes b to int, so the expression is int. Assigning int to byte without cast causes compilation error.

🧠 Conceptual

expert

Type promotion in mixed arithmetic expressions

Given the following Java code, what is the type of the variable 'result' after evaluation?

Java

public class Main {
    public static void main(String[] args) {
        byte b = 1;
        short s = 2;
        char c = 3;
        int i = 4;
        long l = 5L;
        float f = 6.0f;
        double d = 7.0;
        var result = b + s * c - i / l + f - d;
    }
}

A long

B float

C double

D int

💡 Hint

Remember the order of type promotion in mixed expressions: from smaller to larger types.

In mixed arithmetic, all operands are promoted to the largest type involved. Here, double is the largest type, so result is double.