Computer Networking: A Top-Down Approach (7th Edition)
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Author: James Kurose, Keith Ross
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![**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.](https://content.bartleby.com/qna-images/question/6c9872f4-07a5-438e-9711-8d7455ed4876/5d618a33-8b87-4ebb-b0d2-736ba3f115af/wkbrntj.jpeg)
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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