Parsing numeric arguments in Java - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to parse a string into an integer.

Java

int number = Integer.[1]("123");

Drag options to blanks, or click blank then click option'

AparseDouble

BtoString

CvalueOf

DparseInt

Attempts:

3 left

💡 Hint

Common Mistakes

Using toString instead of parseInt causes a compile error.

Using parseDouble returns a double, not an int.

The method parseInt converts a string to an integer.

2fill in blank

medium

Complete the code to parse a string into a double.

Java

double value = Double.[1]("45.67");

Drag options to blanks, or click blank then click option'

AvalueOf

BparseInt

CparseDouble

DtoString

Attempts:

3 left

💡 Hint

Common Mistakes

Using parseInt causes a NumberFormatException for decimal strings.

Using toString returns a string, not a number.

The method parseDouble converts a string to a double value.

3fill in blank

hard

Fix the error in parsing a string to an integer.

Java

int num = Integer.[1]("abc");

Drag options to blanks, or click blank then click option'

AparseDouble

BparseInt

CtoString

DvalueOf

Attempts:

3 left

💡 Hint

Common Mistakes

Using valueOf returns an Integer object, not int primitive.

Using toString is invalid here.

Only parseInt correctly parses strings to int, but "abc" will cause an exception at runtime.

4fill in blank

hard

Fill both blanks to parse a string to a double and assign it to a variable.

Java

Double [1] = Double.[2]("3.14");

Drag options to blanks, or click blank then click option'

Api

BparseDouble

CvalueOf

Dradius

Attempts:

3 left

💡 Hint

Common Mistakes

Using parseDouble returns a primitive double, not a Double object.

Using variable names that don't match the context.

Variable name can be pi. valueOf returns a Double object from a string.

5fill in blank

hard

Fill all three blanks to parse an integer from a string and print it.

Java

String str = "42";
int [1] = Integer.[2](str);
System.out.[3]([1]);

Drag options to blanks, or click blank then click option'

Anumber

BparseInt

Cprintln

Dprint

Attempts:

3 left

💡 Hint

Common Mistakes

Using print instead of println misses the newline.

Using valueOf returns Integer object, not int primitive.

Variable number stores the parsed int. parseInt converts string to int. println prints the value with a new line.

Practice

1.

Which Java method is used to convert a String containing a whole number into an int?

easy

A. Integer.parseInt(String s)

B. Double.parseDouble(String s)

C. String.toInt()

D. Integer.toString(int i)

Solution

1. Step 1: Identify method for integer parsing

   The method Integer.parseInt(String s) converts a string to an int.

2. Step 2: Compare with other options

   Double.parseDouble parses decimals, String.toInt() does not exist, and Integer.toString converts int to string.

3. Final Answer:

   Integer.parseInt(String s) -> Option A

4. Quick Check:

   Whole number parsing = Integer.parseInt [OK]

Hint: Use Integer.parseInt for whole numbers [OK]

Common Mistakes:

• Using Double.parseDouble for integers
• Trying to call non-existent String.toInt()
• Confusing toString() with parsing

2.

Which of the following is the correct syntax to parse a decimal number from a String named input in Java?

double value = ?;

easy

A. Double.parseInt(input);

B. Integer.parseDouble(input);

C. Integer.parseInt(input);

D. Double.parseDouble(input);

Solution

1. Step 1: Identify method for parsing decimals

   Use Double.parseDouble(String s) to convert string to double.

2. Step 2: Check syntax correctness

   Only Double.parseDouble(input) matches correct method and syntax.

3. Final Answer:

   Double.parseDouble(input); -> Option D

4. Quick Check:

   Decimal parsing = Double.parseDouble [OK]

Hint: Use Double.parseDouble for decimals [OK]

Common Mistakes:

• Using Integer.parseInt for decimals
• Calling non-existent Double.parseInt
• Mixing Integer and Double parse methods

3.

What is the output of the following Java code?

String s = "123";
int num = Integer.parseInt(s);
System.out.println(num + 10);

medium

A. 12310

B. 133

C. Error: cannot parse string

D. 10

Solution

1. Step 1: Parse string to integer

   Integer.parseInt("123") converts string "123" to integer 123.

2. Step 2: Add 10 to parsed integer

   Adding 10 to 123 results in 133.

3. Final Answer:

   133 -> Option B

4. Quick Check:

   123 + 10 = 133 [OK]

Hint: Parsed number + 10 means numeric addition [OK]

Common Mistakes:

• Thinking output is string concatenation '12310'
• Assuming parseInt throws error on valid input
• Confusing print output with input string

4.

Identify the error in this Java code snippet that tries to parse a number:

String input = "12a3";
int value = Integer.parseInt(input);
System.out.println(value);

medium

A. Syntax error: missing semicolon

B. No error, prints 123

C. Throws NumberFormatException at runtime

D. Compiles but prints 0

Solution

1. Step 1: Analyze input string content

   The string "12a3" contains a non-digit character 'a', invalid for integer parsing.

2. Step 2: Understand Integer.parseInt behavior

   Parsing invalid numeric strings throws NumberFormatException at runtime.

3. Final Answer:

   Throws NumberFormatException at runtime -> Option C

4. Quick Check:

   Invalid numeric string causes NumberFormatException [OK]

Hint: Invalid chars cause NumberFormatException [OK]

Common Mistakes:

• Assuming parseInt ignores invalid chars
• Expecting compile-time error instead of runtime
• Thinking output will be 0 on error

5.

You want to safely parse a String input to an int in Java, returning -1 if the input is not a valid number. Which code snippet correctly implements this?

hard

A. int parseSafe(String s) {
try {
return Integer.parseInt(s);
} catch (NumberFormatException e) {
return -1;
}
}

B. int parseSafe(String s) {
return Integer.parseInt(s);
}

C. int parseSafe(String s) {
if (s.matches("\\d+")) return Integer.parseInt(s);
else return 0;
}

D. int parseSafe(String s) {
return Integer.valueOf(s);
}

Solution

1. Step 1: Understand safe parsing with exception handling

   Using try-catch around Integer.parseInt catches invalid input and returns fallback.

2. Step 2: Check each option's behavior

   The snippet with try-catch returns parsed int or -1 on error. Snippets without proper handling throw exceptions, return 0 instead of -1, or use valueOf which returns an Integer object not int.

3. Final Answer:

   The try-catch code returning -1 on NumberFormatException -> Option A

4. Quick Check:

   Try-catch with -1 fallback handles invalid input [OK]

Hint: Use try-catch to handle invalid input safely [OK]

Common Mistakes:

• Not handling NumberFormatException
• Returning wrong fallback value
• Confusing Integer.valueOf with parseInt