String type behavior in Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the basic type used to represent text in TypeScript?

The string type is used to represent text in TypeScript. It holds sequences of characters.

beginner

Are strings in TypeScript mutable or immutable?

Strings in TypeScript are immutable. Once created, their content cannot be changed directly.

beginner

How can you create a multi-line string in TypeScript?

Use template literals with backticks (``) to create multi-line strings easily.

beginner

What happens when you use the + operator with strings in TypeScript?

The + operator concatenates (joins) two or more strings into one string.

intermediate

How does TypeScript handle string type checking?

TypeScript enforces that variables declared as string can only hold string values, helping catch errors at compile time.

Which of the following is the correct way to declare a string variable in TypeScript?

A let name: string = 'Alice';

B let name = 123;

C let name: number = 'Alice';

D let name: boolean = true;

Option A correctly declares a variable 'name' with type string and assigns a string value.

What will the following code output?
let greeting = 'Hello';
greeting[0] = 'J';
console.log(greeting);

A Jello

B Error

C Hello

D undefined

Strings are immutable, so changing a character by index does nothing. The original string 'Hello' remains.

How do you create a string that spans multiple lines in TypeScript?

A Using double quotes with \n

B Using backticks (template literals)

C Using single quotes with \n

D Using parentheses

Backticks allow multi-line strings directly without needing escape characters.

What is the result of 'Hello' + ' ' + 'World' in TypeScript?

A 'Hello World'

B Error

C 'Hello+World'

D 'HelloWorld'

The + operator concatenates strings, so it joins them with the space in between.

Which of these is NOT a string method in TypeScript?

A includes()

B toUpperCase()

C slice()

D push()

push() is an array method, not a string method.

Explain how strings behave in TypeScript regarding mutability and concatenation.

Think about whether you can change a letter inside a string directly and how to combine strings.

Describe how TypeScript helps ensure variables hold string values and what happens if you assign a different type.

Consider what TypeScript does before running your code to catch mistakes.

Practice

1. Which of the following is the correct way to declare a string variable in TypeScript?

easy

A. let name: string = "Alice";

B. let name = 123;

C. let name: number = "Alice";

D. let name: boolean = true;

Solution

1. Step 1: Understand TypeScript string type declaration

   In TypeScript, to declare a string variable, use let variableName: string = value; where value is text in quotes.

2. Step 2: Check each option's type and value

   let name: string = "Alice"; uses string type and assigns a text value "Alice" correctly. Others assign wrong types or values.

3. Final Answer:

   let name: string = "Alice"; -> Option A

4. Quick Check:

   Correct string declaration = let name: string = "Alice"; [OK]

Hint: String variables use : string and quotes [OK]

Common Mistakes:

• Assigning number or boolean to string type
• Missing type annotation when required
• Using wrong quotes for strings

2. Which of the following is the correct syntax to create a template string in TypeScript?

easy

A. let greeting = `Hello, ${name}!`;

B. let greeting = Hello, ${name}!;

C. let greeting = "Hello, ${name}!";

D. let greeting = 'Hello, ${name}!';

Solution

1. Step 1: Identify template string syntax

   Template strings use backticks ` and allow embedding variables with ${variable}.

2. Step 2: Check each option's quotes and variable embedding

   let greeting = `Hello, ${name}!`; correctly uses backticks and ${name} syntax. Options A and C use single or double quotes which do not support variable embedding. let greeting = Hello, ${name}!; misses quotes entirely.

3. Final Answer:

   let greeting = `Hello, ${name}!`; -> Option A

4. Quick Check:

   Template strings use backticks = let greeting = `Hello, ${name}!`; [OK]

Hint: Use backticks ` for template strings with ${} [OK]

Common Mistakes:

• Using single or double quotes for templates
• Forgetting backticks
• Not wrapping string in quotes

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

const text: string = "Hello World";
console.log(text.length);

medium

A. 10

B. 11

C. 12

D. Error

Solution

1. Step 1: Understand string length property

   The length property returns the number of characters in the string, including spaces.

2. Step 2: Count characters in "Hello World"

   "Hello World" has 11 characters: 5 letters + 1 space + 5 letters.

3. Final Answer:

   11 -> Option B

4. Quick Check:

   "Hello World" length = 11 [OK]

Hint: Count all characters including spaces for length [OK]

Common Mistakes:

• Ignoring spaces in length count
• Confusing length with last index
• Expecting syntax error

4. Identify the error in this TypeScript code snippet:

let message: string = 'Hello';
message.toUppercase();

medium

A. Strings do not have toUpperCase method

B. No error, outputs 'HELLO'

C. Method name should be toUpperCase() with capital C

D. Missing semicolon after string declaration

Solution

1. Step 1: Check method name spelling

   The correct method to convert string to uppercase is toUpperCase() with capital 'C'.

2. Step 2: Identify error cause

   Using toUppercase() (lowercase 'c') causes a runtime error because the method does not exist.

3. Final Answer:

   Method name should be toUpperCase() with capital C -> Option C

4. Quick Check:

   Correct method spelling = Method name should be toUpperCase() with capital C [OK]

Hint: Remember uppercase C in toUpperCase() method [OK]

Common Mistakes:

• Using wrong case in method names
• Assuming strings lack case methods
• Ignoring runtime errors

5. Given the following code, what does this code check?

const name: string = "Anna";
const reversed = name.split("").reverse().join("");
console.log(reversed === name);

hard

A. If the string length is even

B. If the string is empty

C. If the string contains only letters

D. If the string is a palindrome (same forwards and backwards)

Solution

1. Step 1: Understand string reversal steps

   The code splits the string into characters, reverses the array, then joins back to a string.

2. Step 2: Compare reversed string with original

   It checks if the reversed string equals the original string, which means the string reads the same forwards and backwards.

3. Final Answer:

   If the string is a palindrome (same forwards and backwards) -> Option D

4. Quick Check:

   Reversed equals original means palindrome = If the string is a palindrome (same forwards and backwards) [OK]

Hint: Reversing and comparing checks palindrome [OK]

Common Mistakes:

• Confusing palindrome with empty string
• Thinking it checks string length or letters
• Missing split-reverse-join logic