Birmingham | 26-ITP-May | Gabriel Pawuoi | Sprint 1 | Debugging-Code Reading

debugging/code-reading/readme.md

@@ -17,6 +17,10 @@ Take a look at the following code:
 
 Explain why line 5 and line 8 output different numbers.
 
+## Answer 1
+
+Line 5 outputs `2` and line 8 outputs `1` because of scope. The `x` declared inside `f1` on line 4 is a separate variable that only exists within that function. It shadows the outer `x`. When `f1` finishes, the inner `x` is gone, so line 8 reads the outer `x` which is still `1`.
+
 ## Question 2
 
 Take a look at the following code:
@@ -35,6 +39,16 @@ console.log(y);
 
 What will be the output of this code. Explain your answer in 50 words or less.
 
+## Answer 2
+
+The output is:
+```
+10
+undefined
+ReferenceError: y is not defined
+```
+`f1()` logs `10` (the outer `x`) but returns `undefined` since there is no return statement, so `console.log(f1())` prints `undefined`. Then `console.log(y)` throws a `ReferenceError` because `y` is declared inside `f1` and not accessible outside it.
+
 ## Question 3
 
 Take a look at the following code:
@@ -62,3 +76,12 @@ console.log(y);
 ```
 
 What will be the output of this code. Explain your answer in 50 words or less.
+
+## Answer 3
+
+The output is:
+```
+9
+{ x: 10 }
+```
+Primitives like numbers are passed by value, so `f1` cannot change the original `x`. Objects are passed by reference, so `f2` mutates the original `y` object directly, changing `y.x` from `9` to `10`.