Reference Error
When a variable is called before it is declared. It throws a 'reference error' in that scope.
<pre><code class="lang-plaintext">console.log(a)
//Uncaught ReferenceError: a is not defined.
</code></pre>
This will throw a reference error. As 'a' is not yet been declared yet.
For a error not to be thrown, not only the variable need to be declared but declared in the scope it is being called.
<pre><code class="lang-plaintext">function someFunction(){
 let a = "aman"
 console.log("1", a)
}
console.log("2", a)
</code></pre>
The "1" console will result in "aman".
"2" will result in reference error.
As "a" is defined as "let" which is a block scope i.e. "a" is only accessible inside the "someFunction" scope not outside of it.
"2" console has no access to "a" inside the function, so it will throw reference error.
Type Error
When a certain value is expected by a function but it gets the value of unexpected type due to which the operation could not be performed, then it throws TypeError.
<pre><code class="lang-plaintext">let myAge = 11
console.log(myAge.toUpperCase())
</code></pre>
As "11" is a type = number and "toUppercase()" is a inbuilt function applied only to type "string", the function is expecting a "string" but gets unexpectedly a number so it throws a typeError.
That 11 is not the type it is expecting.
Thanks for reading.

**Reference Error**

When a variable is called before it is declared. It throws a 'reference error' in that scope.

```plaintext
console.log(a)
//Uncaught ReferenceError: a is not defined.
```

This will throw a reference error. As 'a' is not yet been declared yet.

For a error not to be thrown, not only the variable need to be declared but declared in the scope it is being called.

```plaintext
function someFunction(){
 let a = "aman"
 console.log("1", a)
}
console.log("2", a)
```

The "1" console will result in "aman".

"2" will result in reference error.

As "a" is defined as "let" which is a block scope i.e. "a" is only accessible inside the "someFunction" scope not outside of it.

"2" console has no access to "a" inside the function, so it will throw reference error.

**Type Error**

When a certain value is expected by a function but it gets the value of unexpected type due to which the operation could not be performed, then it throws TypeError.

```plaintext
let myAge = 11
console.log(myAge.toUpperCase())
```

As "11" is a type = number and "toUppercase()" is a inbuilt function applied only to type "string", the function is expecting a "string" but gets unexpectedly a number so it throws a typeError.

That 11 is not the type it is expecting.

Thanks for reading.

Reference Error vs TypeError