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.

Practice

1. Which of the following statements about Java type promotion is correct?

easy

A. long is always promoted to byte in expressions.

B. byte, short, and char are promoted to int before arithmetic operations.

C. double is promoted to float when mixed with float.

D. int is promoted to byte automatically in expressions.

Solution

1. Step 1: Understand type promotion basics

   Java promotes smaller integer types (byte, short, char) to int before arithmetic to avoid data loss.

2. Step 2: Evaluate each option

   byte, short, and char are promoted to int before arithmetic operations. correctly states this rule. Options A, B, and C are incorrect because long is not promoted to byte, int is not promoted to byte, and double is not promoted to float.

3. Final Answer:

   byte, short, and char are promoted to int before arithmetic operations. -> Option B

4. Quick Check:

   byte/short/char -> int promotion = D [OK]

Hint: Remember: small integers become int before math [OK]

Common Mistakes:

• Thinking int is promoted to byte
• Confusing float and double promotion
• Assuming long is promoted to smaller types

2. Which of the following Java code snippets correctly compiles without explicit casting due to type promotion?

easy

A. int result = (byte)10 + (short)20;

B. short s = 10 + 20;

C. byte b = 10 + 20;

D. int sum = 10 + 20;

Solution

1. Step 1: Analyze type promotion in arithmetic

   In expressions, byte and short are promoted to int, so 10 + 20 is int.

2. Step 2: Check each assignment

   int result = (byte)10 + (short)20; casts explicitly, so compiles but casting is manual. byte b = 10 + 20; tries to assign int result to byte without cast - error. int sum = 10 + 20; assigns int sum correctly. short s = 10 + 20; tries to assign int to short without cast - error.

3. Final Answer:

   int sum = 10 + 20; -> Option D

4. Quick Check:

   byte/short -> int, assign to int = B [OK]

Hint: Assign promoted int results to int variables [OK]

Common Mistakes:

• Assigning int expression to byte or short without cast
• Forgetting byte/short promote to int
• Assuming automatic narrowing conversion

3. What is the output of the following Java code?

byte b = 10;
short s = 20;
char c = 30;
int result = b + s + c;
System.out.println(result);

medium

A. 60.0

B. Syntax error

C. 60

D. Runtime error

Solution

1. Step 1: Understand type promotion in expression

   byte, short, and char are promoted to int before addition, so b + s + c is int.

2. Step 2: Calculate the sum

   10 + 20 + 30 = 60, assigned to int variable result, then printed.

3. Final Answer:

   60 -> Option C

4. Quick Check:

   byte+short+char -> int sum = 60 [OK]

Hint: Sum small types promoted to int, print int result [OK]

Common Mistakes:

• Expecting byte or short result type
• Confusing output with float/double
• Thinking code causes error

4. Identify the error in the following Java code and fix it:

byte b = 5;
byte c = 10;
byte sum = b + c;
System.out.println(sum);

medium

A. Add explicit cast: byte sum = (byte)(b + c);

B. Change sum to int: int sum = b + c;

C. Change b and c to int: int b = 5, c = 10;

D. No error, code compiles fine.

Solution

1. Step 1: Understand type promotion in addition

   byte variables b and c are promoted to int during addition, so b + c is int.

2. Step 2: Fix assignment error

   Assigning int result to byte sum causes error. Adding explicit cast (byte)(b + c) fixes it.

3. Final Answer:

   Add explicit cast: byte sum = (byte)(b + c); -> Option A

4. Quick Check:

   byte+byte -> int, cast needed for byte = A [OK]

Hint: Cast int result back to byte to fix assignment error [OK]

Common Mistakes:

• Ignoring need for cast when assigning to byte
• Assuming automatic narrowing conversion
• Changing variable types unnecessarily

5. Consider the following code snippet:

float f = 5.5f;
double d = 10.5;
var result = f + d;
System.out.println(((Object)result).getClass().getSimpleName());

What will be printed and why?

hard

A. double because float is promoted to double in mixed expressions

B. float because float is bigger than double

C. float because var infers float type

D. Compilation error due to var usage

Solution

1. Step 1: Understand promotion between float and double

   In expressions mixing float and double, float is promoted to double.

2. Step 2: Determine result type and output

   Variable result is inferred as double by var. Printing class name outputs "Double".

3. Final Answer:

   double because float is promoted to double in mixed expressions -> Option A

4. Quick Check:

   float + double -> double result = A [OK]

Hint: float + double results in double type [OK]

Common Mistakes:

• Assuming var infers float
• Thinking float is bigger than double
• Expecting compilation error with var