a. Write the print outputs at lines 37, 40, 43, and 49-51. b. Identify the pass-by-reference effect in the code and explain the role of the copy() method.

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Subject: Programming language 

**Parameter Passing in Python and Ruby**

The parameter passing method of Python and Ruby is known as **pass-by-assignment** since all data values are objects. In pass-by-assignment, the actual parameter value is assigned to the formal parameter. Therefore, pass-by-assignment effectively behaves like pass-by-reference. 

The following Python code demonstrates this technique and the effect of the assignment operator:

```python
34. x = [5, 1, 6, 9]
35. print(y)
36. y = x
37. y[2] = 0
38. print(x)
39. x[1] = 77
40. print(y)
41. z = x.copy()
42. x[0] = 11
43. print(z)
44. print(x)
45. print(y)
```

**Tasks:**

a. **Write the print outputs at lines 37, 40, 43, and 49-51.**

   - At line 37: `print(y)` outputs nothing initially as `y` hasn't been defined yet before being assigned on line 36.
   - At line 38: `print(x)` outputs `[5, 1, 0, 9]` because `y[2]` is set to `0` affecting `x`.
   - At line 40: `print(y)` outputs `[5, 77, 0, 9]` because `x[1]` is set to `77` affecting `y`.

b. **Identify the pass-by-reference effect in the code and explain the role of the `copy()` method.**

   - The pass-by-reference effect is visible when both `x` and `y` are modified through each other because they reference the same list object initially.
   - The `copy()` method is used to create a shallow copy of `x`, assigning it to `z` so further changes to `x` do not affect `z`. This is why `print(z)` at line 43 outputs `[5, 77, 0, 9]` while `print(x)` outputs `[11, 77, 0, 9]` at line 44 after setting `x[0] = 11`.

This demonstration shows how changes to a mutable object, like a list, affect all references to that object unless a distinct copy is made.
Transcribed Image Text:**Parameter Passing in Python and Ruby** The parameter passing method of Python and Ruby is known as **pass-by-assignment** since all data values are objects. In pass-by-assignment, the actual parameter value is assigned to the formal parameter. Therefore, pass-by-assignment effectively behaves like pass-by-reference. The following Python code demonstrates this technique and the effect of the assignment operator: ```python 34. x = [5, 1, 6, 9] 35. print(y) 36. y = x 37. y[2] = 0 38. print(x) 39. x[1] = 77 40. print(y) 41. z = x.copy() 42. x[0] = 11 43. print(z) 44. print(x) 45. print(y) ``` **Tasks:** a. **Write the print outputs at lines 37, 40, 43, and 49-51.** - At line 37: `print(y)` outputs nothing initially as `y` hasn't been defined yet before being assigned on line 36. - At line 38: `print(x)` outputs `[5, 1, 0, 9]` because `y[2]` is set to `0` affecting `x`. - At line 40: `print(y)` outputs `[5, 77, 0, 9]` because `x[1]` is set to `77` affecting `y`. b. **Identify the pass-by-reference effect in the code and explain the role of the `copy()` method.** - The pass-by-reference effect is visible when both `x` and `y` are modified through each other because they reference the same list object initially. - The `copy()` method is used to create a shallow copy of `x`, assigning it to `z` so further changes to `x` do not affect `z`. This is why `print(z)` at line 43 outputs `[5, 77, 0, 9]` while `print(x)` outputs `[11, 77, 0, 9]` at line 44 after setting `x[0] = 11`. This demonstration shows how changes to a mutable object, like a list, affect all references to that object unless a distinct copy is made.
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