Lesson 1.5: Casting and Range of Variables | AP CSA

Unit 1 · Lesson 1.5 · Code Mechanics

Lesson 1.5: Casting and Range of Variables

Reading time: 10–13 min · Practice: 8 exercises · Mastery: applied scenario

What you'll learn in this lesson

  • 1.5.A. Use (int) and (double) casts to convert between types, predict the result of truncation, and recognize when widening happens automatically.
  • 1.5.A. Apply the rounding formula (int)(x + 0.5) for non-negative doubles.
  • 1.5.B. Explain integer overflow using Integer.MAX_VALUE and Integer.MIN_VALUE, and describe what happens when an int expression exceeds the allowed range.
  • 1.5.C. Describe how double round-off error occurs when a value is more precise than memory allows.

This lesson is part of the AP CSA Course and Exam Description (effective May 2027).

AP exam weight: Casting questions appear on nearly every exam. The most common trap is cast-scope: (int) 2.9 + 1 vs (int)(2.9 + 1) produce different results because the cast only applies to the immediately following operand.

Key Vocabulary

Term Definition
cast operator A type name in parentheses, e.g. (int) or (double), that forces a type conversion.
truncation Dropping all digits after the decimal point when casting double to int; 9.9 becomes 9.
widening Automatic promotion from int to double; no cast required.
Integer.MAX_VALUE The largest value an int can hold: 2,147,483,647 (2^31 - 1).
Integer.MIN_VALUE The smallest value an int can hold: -2,147,483,648 (-2^31).
integer overflow When an int expression produces a value outside the allowed range; the value wraps around silently.
round-off error Imprecision in double values because some decimals cannot be represented exactly in binary floating-point.
rounding formula (int)(x + 0.5) rounds a non-negative double to the nearest integer by truncating after adding 0.5.

Casting: converting between primitive types

A cast forces a value from one numeric type to another. Java provides two casting operators for primitives: (int) and (double).

The Two Cast Operators

(int) expression — converts a double to an int by dropping all digits after the decimal point (truncation, not rounding).
(double) expression — converts an int to a double, adding a decimal point. No information is lost.

Examples

double d = 9.99;
int n = (int) d;         // n = 9 (truncated, not rounded)

int x = 5;
double r = (double) x;  // r = 5.0

double avg = (double) 7 / 2;  // 7 cast to 7.0, then 7.0/2 = 3.5
double bad = (double)(7 / 2); // 7/2 = 3 first (int div), then cast to 3.0

AP Trap: cast scope

A cast operator applies only to the immediately following operand, not the whole expression. (int) 2.9 + 1.6 casts only 2.9 to 2, then adds 1.6, giving 3.6. To cast the entire sum, use parentheses: (int)(2.9 + 1.6) gives 4.

Widening: automatic int-to-double promotion

When an int value is used where a double is expected, Java automatically widens it. This happens without a cast whenever: you assign an int to a double variable, or you use an int and a double together in an arithmetic expression.

Widening is Automatic; Narrowing Requires a Cast

double d = 5; — valid, 5 widens to 5.0.
int n = 5.0; — compile error, narrowing requires explicit (int) cast.

Rounding a double to the nearest integer

Because (int) truncates toward zero, rounding requires adding 0.5 before casting.

Rounding Formula (non-negative values)

int rounded = (int)(x + 0.5);

Examples: x = 3.2 → 3.2 + 0.5 = 3.7 → truncates to 3. x = 3.7 → 4.2 → truncates to 4.

Integer range and overflow

Java uses exactly 4 bytes (32 bits) to store an int. This limits the allowed range.

The Constants

Integer.MAX_VALUE = 2,147,483,647 (231 − 1)
Integer.MIN_VALUE = −2,147,483,648 (−231)

If an arithmetic expression would produce a value outside this range, integer overflow occurs. Java does not throw an exception. Instead, the value wraps around to the other end of the range.

Overflow Wraps, Not Crashes

int big = Integer.MAX_VALUE;
big = big + 1;   // overflows to Integer.MIN_VALUE (-2147483648)

The program keeps running with the wrong value. No exception is thrown. This is one reason to use long when dealing with very large numbers in real programs—but long is outside the scope of the AP exam.

Double round-off error

Computers store double values in a fixed amount of memory (8 bytes). Some decimal values cannot be represented exactly in binary floating point. When a value is more precise than the memory can hold, a round-off error occurs: the stored value is the nearest representable number.

Classic Example

System.out.println(0.1 + 0.2); // prints 0.30000000000000004

To minimize round-off errors, prefer int arithmetic whenever exact results are required (for example, work in cents rather than dollars).

AP Scope Note

Other decimal types that avoid round-off errors (such as BigDecimal) are outside the scope of the AP exam. You need to know that round-off error exists and when it occurs, but you will not be asked to use alternative types.

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Tier 2 · AP Practice

Practice Questions

What value is stored in result after the following statement executes?

double result = (double)(5 / 2) + (double) 1 / 4;
Evaluate each cast and each division carefully before choosing. The cast on the left applies to the whole sub-expression; the cast on the right applies only to 1.
A. 2.75
B. 3.0
C. 2.5
D. 2.25
D. Left side: 5 / 2 is int division = 2; then (double) 2 = 2.0. Right side: (double) 1 = 1.0; then 1.0 / 4 = 0.25 (double / int = double). Sum: 2.0 + 0.25 = 2.25. A would result if 5/2 gave 2.5. B would result if both sides rounded up. C uses 5/2 = 2.5 incorrectly.
Consider the following code segment.

double d = 7.9;
int trunc = (int) d;
int rounded = (int)(d + 0.5);
System.out.println(trunc + " " + rounded);
Predict both values before reading the options. What does truncation give? What does the rounding formula give?
A. 8 8
B. 7 8
C. 7 7
D. 8 9
B. (int) 7.9 truncates to 7. 7.9 + 0.5 = 8.4; (int) 8.4 truncates to 8. Prints 7 8. A is wrong because truncation drops the decimal, it does not round. C applies truncation to both. D adds an extra 1 incorrectly.
Which of the following expressions evaluates to 3.5?

I. (double) 7 / 2
II. (double)(7 / 2)
III. 7 / (double) 2
A. I and III only
B. II only
C. I, II, and III
D. I only
A. Expression I: cast applies to 7 first, giving 7.0. Then 7.0 / 2 = 3.5 (double / int). Expression II: 7 / 2 = 3 first (int division), then cast gives 3.0—not 3.5. Expression III: cast applies to 2, giving 2.0. Then 7 / 2.0 = 3.5 (int / double). I and III both produce 3.5; II produces 3.0.
A student writes the following code.

int n = Integer.MAX_VALUE;
n = n + 1;
System.out.println(n);

Which of the following best describes what happens?
A. A compile error occurs because adding 1 to Integer.MAX_VALUE is not allowed.
B. An ArithmeticException is thrown at run time because the value exceeds the int range.
C. The program runs and prints Integer.MIN_VALUE because integer overflow wraps the value to the minimum.
D. The program runs and prints 2147483648 because Java automatically promotes n to long.
C. Integer overflow silently wraps the value. Adding 1 to Integer.MAX_VALUE (2,147,483,647) wraps around to Integer.MIN_VALUE (−2,147,483,648). No exception is thrown and no error is reported. A is wrong: the compiler allows this. B is wrong: overflow does not throw an exception for integer types. D is wrong: Java does not auto-promote int to long on overflow.
Which of the following statements will cause a compile error?

I. double d = 4;
II. int n = 4.0;
III. int m = (int) 4.0;
A. I only
B. II only
C. I and III only
D. II and III only
B. Statement I: assigning int 4 to double is widening—valid. Statement II: assigning double 4.0 to int is narrowing without a cast—compile error. Statement III: (int) 4.0 explicitly casts 4.0 to 4, then assigns to int—valid. Only II fails.
A student expects the following code to print 0.3 but gets a different result.

double x = 0.1 + 0.2;
System.out.println(x);

Which of the following best explains the unexpected output?
A. The + operator does not work correctly with double values; Math.add should be used instead.
B. 0.1 and 0.2 are stored as int values internally, so their sum truncates.
C. The variable x should be declared as float rather than double to get exact decimal arithmetic.
D. 0.1 and 0.2 cannot be represented exactly in binary floating point, so their stored values differ slightly from the true decimal values, producing a round-off error in the sum.
D. Binary floating-point representation cannot exactly represent most decimal fractions. 0.1 is stored as the nearest representable binary value, which is slightly more than 0.1. The same is true for 0.2. Their sum is slightly more than 0.3, producing output like 0.30000000000000004. A is wrong: + works correctly with doubles. B is wrong: they are stored as doubles, not ints. C is wrong: float also has round-off error and is less precise.
Consider the following three statements about integer overflow in Java.

I. Integer overflow throws an ArithmeticException at run time.
II. If Integer.MAX_VALUE is incremented by 1, the result equals Integer.MIN_VALUE.
III. Java automatically promotes an int variable to long when overflow would occur.

Which of the statements above are TRUE?
A. I only
B. I and II only
C. II only
D. II and III only
C. Statement I is FALSE: integer overflow does not throw an exception. The value wraps silently. Statement II is TRUE: Integer.MAX_VALUE + 1 wraps to Integer.MIN_VALUE due to binary arithmetic. Statement III is FALSE: Java does not automatically promote to long on overflow. The variable stays int with the wrapped value.
What value is stored in result after the following statement executes?

int result = (int)(3.6 + 2.9);
Evaluate inside the parentheses first, then apply the cast.
A. 6
B. 5
C. 7
D. 6.5
A. 3.6 + 2.9 = 6.5 (double arithmetic inside the parentheses). Then (int) 6.5 truncates to 6. B would result from casting each value individually and adding: (int)3.6 + (int)2.9 = 3 + 2 = 5. C would result from rounding up. D is wrong because the cast to int produces an integer.
PRACTICE WITH A GAME — CHOOSE ONE:

Output Predictor — Casting

Predict the exact printed value after each cast or conversion.

Question 1 of 6 Score: 0 

        

          
          
        

        
      

      
    

  

  

    

      

        
Bug Hunt — Casting

        
Each snippet has exactly one buggy line. Click it, then submit.

      

      

        Bug 1 of 7
        Caught: 0
      

      

        

        

        
        
        
      

      
    

  







  Tier 3 · AP Mastery Challenge
  
The Track Team Timer

  
A student writes a program that reads a runner's time in seconds (as a double) and computes the whole seconds and the tenths of a second separately for display. A classmate tests it and finds the output is wrong.

  
double rawTime = 12.7;
int wholeSeconds = (int) rawTime;
int tenths = (int)((rawTime - wholeSeconds) * 10);
System.out.println(wholeSeconds + "." + tenths);


  
Part A: Predict the output

  

    
What does the program print for rawTime = 12.7?

    
Trace each line. What is wholeSeconds? What is rawTime - wholeSeconds? Then what is tenths?

    

      

A. 12.8

      

B. 12.7

      

C. 12.0

      

D. 12.6

    

    
    

      D. wholeSeconds = (int) 12.7 = 12. rawTime - wholeSeconds = 12.7 - 12 = 0.7 in theory, but due to double round-off error the stored value of 0.7 is slightly less than 0.7 (approximately 0.699999...). Multiplying by 10 gives approximately 6.9999..., and (int) truncates that to 6. So tenths = 6 and the program prints 12.6. This is the round-off error trap.
    

  


  
Part B: Identify the fix

  

    
Which of the following changes to line 3 would most reliably fix the output so tenths correctly stores 7?

    

      

A. int tenths = (int)((rawTime - wholeSeconds) * 10 - 0.5);


      

B. int tenths = (int)((rawTime - wholeSeconds) * 10 + 0.5);


      

C. int tenths = (int) rawTime % 10;


      

D. int tenths = (int)(rawTime * 10) - wholeSeconds;


    

    
    

      B. Adding 0.5 before truncating is the standard rounding fix for non-negative values. 0.6999... * 10 = 6.999...; adding 0.5 gives 7.499..., which truncates to 7. A subtracts 0.5, making the problem worse. C extracts the tens digit of the truncated integer (which is wrong). D: (int)(rawTime * 10) = (int)(127.0) = 127, then 127 - 12 = 115—completely wrong.
    

  


  
Part C: Structured response

  
A teammate suggests storing money as a double (e.g., double price = 0.10) and adding values like 0.10 + 0.20 to compute totals. In three to five sentences, explain why this approach risks incorrect results, what Java concept causes the problem, and what strategy avoids it.

  
  





  Extension · Beyond the Exam
  
BigDecimal for exact decimal arithmetic

  
java.math.BigDecimal stores decimal values exactly using arbitrary-precision arithmetic. new BigDecimal("0.1").add(new BigDecimal("0.2")) returns exactly 0.3. It is slower than double and more verbose, but it is the right tool whenever exact decimal results matter—financial calculations, tax computations, currency conversions. It is outside the AP exam scope but worth knowing exists.

















  

















  
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