11.2. print

In Section 9.1 we defined a structure named Time and wrote a function named printTime

struct Time {
  int hour, minute;
  double second;
};

void printTime (const Time& time) {
  cout << time.hour << ":" << time.minute << ":" << time.second << endl;
}

To call this function, we had to pass a Time object as a parameter.

Time currentTime = { 9, 14, 30.0 };
printTime (currentTime);

To make printTime into a member function, the first step is to change the name of the function from printTime to Time::print. The :: operator separates the name of the structure from the name of the function; together they indicate that this is a function named print that can be invoked on a Time structure.

The next step is to eliminate the parameter. Instead of passing an object as an argument, we are going to invoke the function on an object.

As a result, inside the function, we no longer have a parameter named time. Instead, we have a current object, which is the object the function is invoked on. We can refer to the current object using the C++ keyword this.

One thing that makes life a little difficult is that this is actually a pointer to a structure, rather than a structure itself. A pointer is similar to a reference, but I don’t want to go into the details of using pointers yet. The only pointer operation we need for now is the * operator, which converts a structure pointer into a structure. In the following function, we use it to assign the value of this to a local variable named time.

void Time::print () {
  Time time = *this;
  cout << time.hour << ":" << time.minute << ":" << time.second << endl;
}

The first two lines of this function changed quite a bit as we transformed it into a member function, but notice that the output statement itself did not change at all.

In order to invoke the new version of print, we have to invoke it on a Time object:

Time currentTime = { 9, 14, 30.0 };
currentTime.print ();

The last step of the transformation process is that we have to declare the new function inside the structure definition:

struct Time {
  int hour, minute;
  double second;

  void print ();
};

Note

You should only use the scope resolution operator :: if you define a member function outside of its structure definition. If you define a function inside of the structure definition, you define it as you would any other function.

A function declaration looks just like the first line of the function definition, except that it has a semi-colon at the end. The declaration describes the interface of the function; that is, the number and types of the arguments, and the type of the return value.

When you declare a function, you are making a promise to the compiler that you will, at some point later on in the program, provide a definition for the function. This definition is sometimes called the implementation of the function, since it contains the details of how the function works. If you omit the definition, or provide a definition that has an interface different from what you promised, the compiler will complain.

Feel free to mess around with input for currentTime in the active code below!

#include <iostream> using namespace std; struct Time { int hour, minute; double second; void print (); }; int main() { Time currentTime = { 9, 14, 30.0 }; currentTime.print (); } ==== void Time::print () { cout << hour << ":" << minute << ":" << second << endl; }

Q-3: We have a free-standing function called dog_bark which takes a Dog object as a parameter. What step(s) do we need to take to convert dog_bark(const Dog& dog) to a member function of the Dog class?

• Change the name of the function to Dog::bark
• Incorrect! You don't need to rename the function unless you define it outside of the structure definition.
• Remove the Dog parameter
• Correct! We no longer need to pass a Dog as an argument, since we are going to be invoking the function on a Dog object.
• Operate on the current Dog object by using *this
• Correct! To get the current object, we need to dereference the this pointer using *.
• Declare the function inside of the Dog structure definition
• Correct! Member functions are declared inside of structure definitions.

Create the Dog object with member functions bark and is_teacup_dog (if the weight of the dog is less than 4 pounds) Write the functions in the same order they appear inside the structure.

        struct Dog {
---
 int age, weight;
 string breed;
---
 void bark();
 bool is_teacup_dog();
---
};
---
}                         #paired
---
void Dog::bark() {
---
void bark() {                         #paired
---
 cout << "RUFF!" << endl;
}
---
bool Dog::is_teacup_dog() {
---
bool is_teacup_dog() {                          #paired
---
 Dog dog = *this;                          #distractor
---
 if (weight < 4) {
   return true;
 }
 return false;
}
        

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