Variables and Primitive Types in Java: Declarations, Conversions and Worked Examples

int score = 87; looks easy, but Java output questions quickly mix literal types, integer division, numeric promotion, narrowing casts and overflow. A variable's declared type limits the values it can hold and controls how its expressions are evaluated. The Coding & DSA Courses for Placements page connects these Java foundations to the wider placement-preparation route.

Related reading: Java strings and the string pool and Java conditionals.

1. Java variables: declaration, initialisation and assignment

In int score = 87;, int is the declared type, score is the identifier and 87 is the initializer. You can separate those steps:

int attempts;
attempts = 3;

The second statement assigns a value without redeclaring attempts or changing its type.

studentScore, _index2 and $limit are legal names. 2score cannot start with a digit, and class is a keyword. Java is case-sensitive, so score and Score differ. Although $ is legal, it is normally left to generated or framework-oriented names.

A local variable such as int count; must be definitely assigned before it is read. An instance field declared as int count; receives the default value 0. A declaration such as final int MAX_ATTEMPTS = 3; allows one assignment and no reassignment, but final alone does not make every variable a compile-time constant.

2. Java primitive types: all eight value sets and ranges

Java has exactly eight primitive types:

Both floating-point types also represent negative values, signed zero, infinities and NaN. Java does not define a programmer-visible storage size for boolean, so treating it as a one-byte type is incorrect.

Use int for ordinary whole-number arithmetic, long for values beyond the int range, and double for general fractional work. Use char for one UTF-16 code unit and boolean for a condition. The powers-of-two boundaries behind signed integer ranges become clearer after Number Systems and Base Conversions Explained.

3. Java literals: suffixes, bases, characters and underscores

The literal 87 has type int. 4_294_967_296L needs L because its value exceeds the int range, and uppercase L is clearer than lowercase l. Decimal floating literals are double by default, so 91.5F needs F, while 87.4 does not.

These declarations all store 42:

int decimal = 42;
int binary = 0b101010;
int hex = 0x2A;

Underscores make 1_000_000 readable, but cannot sit at either end or beside a radix prefix, decimal point or suffix. 'A' is a char; "A" is a String. char omega = '\u03A9'; stores the Greek capital omega, and char letter = 65; stores 'A' because 65 is an in-range constant expression. Some supplementary Unicode code points need a surrogate pair, so one char does not guarantee one complete visible character.

4. Java primitive types: a complete worked program trace

public class PrimitiveReport {
    public static void main(String[] args) {
        byte subjects = 5;
        short maxPerSubject = 100;
        int total = 437;
        long studentId = 4_294_967_296L;
        float attendance = 91.5F;

        double average = total / (double) subjects;
        char grade = average >= 80.0 ? 'A' : 'B';
        boolean passed = total >= subjects * 40
                && attendance >= 75.0F;

        System.out.println(studentId);
        System.out.println(subjects * maxPerSubject);
        System.out.println(average);
        System.out.println(grade);
        System.out.println(passed);
    }
}

The cast changes 5 to 5.0, so 437 / 5.0 = 87.4. Then 87.4 >= 80.0 selects 'A'. For the pass condition, the byte value 5 is promoted to int: 5 * 40 = 200. Both 437 >= 200 and 91.5F >= 75.0F are true, so true && true produces true.

The output is:

4294967296
500
87.4
A
true

subjects * maxPerSubject equals 500 and has type int after binary numeric promotion, although both stored operands use smaller integral types.

5. Java type conversion and numeric promotion

Widening preserves byte small = 120; int wide = small; as 120. Narrowing can change a value: int n = 260; byte wrapped = (byte) n; keeps the low eight bits and produces 4. double mark = 87.9; int whole = (int) mark; produces 87 because conversion discards the fractional part toward zero.

With byte a = 100; byte b = 27;, a + b has type int and value 127. Therefore int sum = a + b; compiles, but byte sum = a + b; does not. byte constant = 100 + 27; does compile because the compiler knows that the constant value 127 fits in byte.

char letter = 'A'; int code = letter; gives 65. char next = (char) (letter + 2); gives 'C'. A boolean never converts to or from an integer in Java, so C-style 0 and 1 truth values are invalid.

6. Java primitive-type traps: overflow, division and precision

Convert an operand before division when you need a fraction. Use Math.addExact when silent integer wraparound is unacceptable. For the representation behind the last two rows, read Floating Point Representation: IEEE 754 Format. Exact decimal money arithmetic normally needs BigDecimal, which is a class, not a primitive type.

7. How questions test variables and primitive types in Java

Common questions ask you to identify a literal's type, decide whether an assignment compiles, predict promotion and casts, trace ++ or compound assignment, and spot overflow or lost precision. Four quick checks expose the main rules:

• double x = 9 / 4; gives 2.0.

• double y = 9 / 4.0; gives 2.25.

• After byte b = 127; b++;, b is -128 because increment includes an implicit narrowing conversion.

• After char ch = 'A'; ch += 2;, ch is 'C'. In contrast, ch = ch + 2; needs an explicit cast because ch + 2 is an int.

Solve these in three steps: annotate each operand's type, apply promotion and the operator, then apply any cast or destination conversion.

8. Variables and primitive types in Java: the short version

Keep six rules together:

1. Distinguish declaration from later assignment.

2. Know the value sets of all eight primitive types.

3. Read literal suffixes and spellings before calculating.

4. Promote operands before evaluating an expression.

5. Treat narrowing as a possible value change.

6. Test overflow and floating precision instead of assuming arithmetic is exact.

The Java Course: Concepts, MCQs & Coding Questions continues from primitive fundamentals into Java concepts, practice questions and coding work.

Finish with this trace before compiling: byte a = 120; int b = a + 140; byte c = (byte) b; double d = b / 6; double e = b / 6.0; Predict a, b, c, d and e using the three-step method.

Pause, then check: the values are 120, 260, 4, 43.0 and 43.333333333333336.