Global augmentation in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add a new property greet to the global Window interface.

Typescript

declare global {
  interface Window {
    greet: () => string;
  }
}

window.[1] = () => "Hello!";

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

Ahello

Bgreet

CsayHello

Dwelcome

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different property name than declared in the interface.

Forgetting to assign the function to the window object.

We add the greet property to the Window interface, so we must assign the function to window.greet.

2fill in blank

medium

Complete the code to add a new method double to the global Array interface that returns an array with all numbers doubled.

Typescript

declare global {
  interface Array<T> {
    double(): number[];
  }
}

Array.prototype.[1] = function() {
  return this.map(x => x * 2);
};

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

Adouble

Brepeat

CtimesTwo

Dmultiply

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different method name than declared.

Not adding the method to Array.prototype.

The method name must match the one declared in the global interface, which is double.

3fill in blank

hard

Fix the error in the global augmentation by completing the code to add a new property version of type string to the global Navigator interface.

Typescript

declare global {
  interface Navigator {
    [1]: string;
  }
}

console.log(navigator.version);

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

AnavVersion

Bver

CversionNumber

Dversion

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different property name than accessed.

Misspelling the property name.

The property name must be version to match the usage navigator.version.

4fill in blank

hard

Fill both blanks to add a new method toTitleCase to the global String interface that capitalizes the first letter of each word.

Typescript

declare global {
  interface String {
    [1](): string;
  }
}

String.prototype.[2] = function() {
  return this.split(' ').map(word => word.charAt(0).toUpperCase() + word.slice(1)).join(' ');
};

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

AtoTitleCase

BcapitalizeWords

CtitleCase

DtoUpperWords

Attempts:

3 left

💡 Hint

Common Mistakes

Using different names in interface and prototype.

Misspelling the method name.

The method name must be toTitleCase both in the interface and on the prototype.

5fill in blank

hard

Fill all three blanks to add a global augmentation for Document interface with a method getElementByDataId that returns an element by its data-id attribute.

Typescript

declare global {
  interface Document {
    [1](id: string): HTMLElement | null;
  }
}

Document.prototype.[2] = function(id: string) {
  return this.querySelector('[[3]="' + id + '"]');
};

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

AgetElementByDataId

Bdata-id

CgetDataElement

DdataId

Attempts:

3 left

💡 Hint

Common Mistakes

Using different method names in interface and prototype.

Incorrect attribute name in selector.

The method name is getElementByDataId in interface and prototype. The attribute selector uses data-id.

Practice

1. What is the main purpose of declare global in TypeScript?

easy

A. To import external modules into the current file

B. To declare a new local variable inside a function

C. To safely add or modify types on global objects or modules

D. To create a new class with global scope

Solution

1. Step 1: Understand the role of declare global

   This keyword is used inside a module to extend or modify global types safely without changing original files.

2. Step 2: Compare with other options

   Options B, C, and D describe unrelated TypeScript features like local variables, imports, or classes.

3. Final Answer:

   To safely add or modify types on global objects or modules -> Option C

4. Quick Check:

   declare global = add/modify global types [OK]

Hint: Remember: declare global extends global types safely [OK]

Common Mistakes:

• Confusing declare global with import statements
• Thinking it declares local variables
• Assuming it creates new classes

2. Which syntax correctly starts a global augmentation block inside a module in TypeScript?

easy

A. declare global { interface Window { customProp: string; } }

B. global declare { interface Window { customProp: string; } }

C. declare module global { interface Window { customProp: string; } }

D. module global declare { interface Window { customProp: string; } }

Solution

1. Step 1: Identify correct syntax for global augmentation

   The correct syntax uses declare global { ... } inside a module to augment global types.

2. Step 2: Check other options for syntax errors

   Options A, B, and C use incorrect order or keywords that are invalid in TypeScript.

3. Final Answer:

   declare global { interface Window { customProp: string; } } -> Option A

4. Quick Check:

   declare global { ... } is correct syntax [OK]

Hint: Use 'declare global { }' exactly to augment globals [OK]

Common Mistakes:

• Swapping 'declare' and 'global' keywords
• Using 'declare module global' incorrectly
• Missing curly braces after declare global

3. What will be the output of this TypeScript code after global augmentation?

export {};
declare global {
  interface Array<T> {
    first(): T | undefined;
  }
}

Array.prototype.first = function() {
  return this.length > 0 ? this[0] : undefined;
};

const nums = [10, 20, 30];
console.log(nums.first());

medium

A. undefined

B. 30

C. Error: Property 'first' does not exist on type 'number[]'

D. 10

Solution

1. Step 1: Understand the global augmentation of Array interface

   The code adds a new method first() to all arrays returning the first element or undefined if empty.

2. Step 2: Analyze the output of calling nums.first()

   The array nums has elements [10, 20, 30], so first() returns 10, the first element.

3. Final Answer:

   10 -> Option D

4. Quick Check:

   nums.first() = 10 [OK]

Hint: Augmented methods work like normal after declaration [OK]

Common Mistakes:

• Expecting a TypeScript error without module context
• Thinking first() returns undefined always
• Confusing first element with last element

4. Identify the error in this global augmentation code and fix it:

declare global {
  interface String {
    shout(): string;
  }
}

String.prototype.shout = function() {
  return this.toUpperCase() + "!";
};

medium

A. Cannot augment String interface globally

B. Missing export {} to make file a module

C. Incorrect method name, should be 'shoutOut'

D. Prototype assignment must be inside declare global block

Solution

1. Step 1: Check module context requirement for global augmentation

   Global augmentation requires the file to be a module, so it needs an export {} or import statement.

2. Step 2: Identify missing export statement

   The code lacks export {}, so TypeScript treats it as a script, causing errors with declare global.

3. Final Answer:

   Missing export {} to make file a module -> Option B

4. Quick Check:

   File must be a module for declare global [OK]

Hint: Add 'export {}' to enable global augmentation [OK]

Common Mistakes:

• Forgetting to make file a module
• Trying to assign prototype inside declare global
• Renaming methods unnecessarily

5. You want to add a global method logInfo() to the console object that prefixes messages with '[INFO]'. Which is the correct way to augment the global Console interface and implement this method?

hard

A. export {};
declare global {
interface Console {
logInfo(message: string): void;
}
}

console.logInfo = (msg) => console.log(`[INFO] ${msg}`);

B. declare module 'console' {
interface Console {
logInfo(message: string): void;
}
}

console.logInfo = (msg) => console.log(`[INFO] ${msg}`);

C. declare global {
interface Console {
logInfo(message: string): void;
}
}

console.prototype.logInfo = function(msg) { console.log(`[INFO] ${msg}`); };

D. export {};
declare global {
interface Console {
logInfo(message: string): void;
}
}

console.prototype.logInfo = (msg) => console.log(`[INFO] ${msg}`);

Solution

1. Step 1: Correctly augment the global Console interface

   Use export {} to make the file a module, then declare global with interface Console adding logInfo.

2. Step 2: Properly assign the method to console object

   Assign console.logInfo directly, not on console.prototype because console is an object, not a class instance.

3. Final Answer:

   export {}; declare global { interface Console { logInfo(message: string): void; } } console.logInfo = (msg) => console.log(`[INFO] ${msg}`); -> Option A

4. Quick Check:

   Use declare global + direct console method assignment [OK]

Hint: Augment interface then assign method directly on console [OK]

Common Mistakes:

• Using declare module 'console' instead of declare global
• Assigning method to console.prototype (wrong for object)
• Forgetting export {} to make file a module