11. Challenge 5

References

Lower level languages, like C++, give you control over the way your objects are stored and accessed in memory. As a comparison, let's start by looking at how Python hides memory access from you.

def dict_modifier(d, key):
  d.pop(key, None)

sample_dict = {'some_key': 'some value'}
dict_modifier(sample_dict, 'some_key')
print(sample_dict)  # {}

In this case, dict_modifier removes a key from the dictionary passed to it. In effect, it treats the dictionary argument as a reference.

You may want to draw the conclusion that Python functions always treat arguments as references, but that's not true.

def adder(n):
  n += 1

i = 1
adder(i)
print(i)  # 1

Python does not pass primitives as references. As a result, adder receives a copy of i and i remains unchanged outside the function.

To be fair, I think the way Python handles arguments is perfectly reasonable. But what's happening behind the scenes is not perfectly obvious from the syntax alone.

C++, on the other hand, makes you decide how you'd like to treat function arguments. It also gives you control over how you'd like to access objects in memory, whether that's by simply referring to their address in memory or by their actual value.

Pass by Reference

In this last challenge, I want you to finish the doubler function that doubles an int passed to it as a reference. You'll need to finish defining the function's parameters and body.

Start Quiz:

/*
You'll need to finish the Doubler function here AND fix the parameters
of its signature in Doubler.h.
*/

#include <Doubler.h>

void Doubler(/* replace this comment with your code */)
{
    // your code here!
}

void Doubler(/* put the right arguments here too! */);

/*
Like always, no need to change anything here, but feel free to play with the
test case.
*/
#include <iostream>
#include <Doubler.h>

using namespace std;

int main()
{
    int value = 25;
    
    cout << "Original value: " << value << endl;
    
    Doubler(value);
    
    cout << "Doubled value: " << value << endl;
    
    return 0;
}