We
have already learned how groups of sequential data can be used in C++. But this
is somewhat restrictive, since in many occasions what we want to store are not
mere sequences of elements all of the same data type, but sets of different
elements with different data types.
Data structures
A data
structure is a group of data elements grouped together under one name. These
data elements, known as members, can have different types and different
lengths. Data structures are declared in C++ using the following syntax:
struct structure_name {
member_type1 member_name1;
member_type2 member_name2;
member_type3 member_name3;
.
.
} object_names;
member_type1 member_name1;
member_type2 member_name2;
member_type3 member_name3;
.
.
} object_names;
where structure_name is a name for the structure type, object_name can
be a set of valid identifiers for objects that have the type of this structure.
Within braces {
} there is a list with the data members, each one is specified
with a type and a valid identifier as its name.
The first thing we have to
know is that a data structure creates a new type: Once a data structure is
declared, a new type with the identifier specified as structure_name is created and can be used in the rest of the program as if it
was any other type. For example:
struct product { int weight;
float price;
} ;
product apple;
product banana, melon;
|
We have first declared a
structure type called product with two members: weight and price, each of a different
fundamental type. We have then used this name of the structure type (product) to declare three objects of that type: apple, banana and melon as we would have done with any fundamental data type.
Once declared, product has become a new valid type name like the fundamental ones int, char or short and from that point on we are able to declare objects
(variables) of this compound new type, like we have done with apple, banana and melon.
Right at the end of the struct declaration, and before the ending semicolon, we can use the
optional field object_name to directly declare objects of the structure type. For example,
we can also declare the structure objects apple, banana and melon at the moment we define the
data structure type this way:
struct product { int weight;
float price;
} apple, banana, melon;
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It is important to clearly
differentiate between what is the structure type name, and what is an object
(variable) that has this structure type. We can instantiate many objects (i.e.
variables, like apple, banana and melon) from a single structure type (product).
Once we have declared our
three objects of a determined structure type (apple, banana and melon) we can operate directly
with their members. To do that we use a dot (.)
inserted between the object name and the member name. For example, we could
operate with any of these elements as if they were standard variables of their
respective types:
apple.weight
apple.price
banana.weight
banana.price
melon.weight
melon.price
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Each one of these has the
data type corresponding to the member they refer to: apple.weight, banana.weight and melon.weight are of type int, while apple.price, banana.price and melon.price are of type float.
Let's see a real example
where you can see how a structure type can be used in the same way as
fundamental types:
// example about structures #include <iostream> #include <string> #include <sstream> using namespace std;
struct movies_t { string title;
int year;
} mine, yours;
void printmovie (movies_t movie); int main () {
string mystr;
mine.title = "2001 A Space Odyssey";
mine.year = 1968;
cout << "Enter title: ";
getline (cin,yours.title);
cout << "Enter year: ";
getline (cin,mystr);
stringstream(mystr) >> yours.year;
cout << "My favorite movie is:\n ";
printmovie (mine);
cout << "And yours is:\n ";
printmovie (yours);
return 0;
}
void printmovie (movies_t movie) {
cout << movie.title;
cout << " (" << movie.year << ")\n";
}
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Enter title: Alien
Enter year: 1979
My favorite movie is:
2001 A Space Odyssey (1968)
And yours is:
Alien (1979)
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The example shows how we can
use the members of an object as regular variables. For example, the member yours.year is a valid variable of type int, and mine.title is a valid variable of type string.
The objects mine and yours can also be treated as valid
variables of type movies_t, for example we have passed
them to the function printmovie as we would have done with
regular variables. Therefore, one of the most important advantages of data
structures is that we can either refer to their members individually or to the
entire structure as a block with only one identifier.
Data structures are a feature
that can be used to represent databases, especially if we consider the
possibility of building arrays of them:
// array of structures #include <iostream> #include <string> #include <sstream> using namespace std;
#define N_MOVIES 3
struct movies_t { string title;
int year;
} films [N_MOVIES];
void printmovie (movies_t movie);
int main () {
string mystr;
int n;
for (n=0; n<N_MOVIES; n++)
{
cout << "Enter title: ";
getline (cin,films[n].title);
cout << "Enter year: ";
getline (cin,mystr);
stringstream(mystr) >> films[n].year;
}
cout << "\nYou have entered these movies:\n";
for (n=0; n<N_MOVIES; n++)
printmovie (films[n]);
return 0;
}
void printmovie (movies_t movie) {
cout << movie.title;
cout << " (" << movie.year << ")\n";
}
|
Enter title: Blade Runner
Enter year: 1982
Enter title: Matrix
Enter year: 1999
Enter title: Taxi Driver
Enter year: 1976
You have entered these movies:
Blade Runner (1982)
Matrix (1999)
Taxi Driver (1976)
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Pointers to structures
Like
any other type, structures can be pointed by its own type of pointers:
struct movies_t { string title;
int year;
};
movies_t amovie;
movies_t * pmovie;
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Here amovie is an object of structure type movies_t, and pmovie is a pointer to point to objects of structure type movies_t. So, the following code would also be valid:
pmovie = &amovie;
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The value of the pointer pmovie would be assigned to a reference to the object amovie (its memory address).
We will now go with another
example that includes pointers, which will serve to introduce a new operator:
the arrow operator (->):
// pointers to structures #include <iostream> #include <string> #include <sstream> using namespace std;
struct movies_t { string title;
int year;
};
int main () {
string mystr;
movies_t amovie;
movies_t * pmovie;
pmovie = &amovie;
cout << "Enter title: ";
getline (cin, pmovie->title);
cout << "Enter year: ";
getline (cin, mystr);
(stringstream) mystr >> pmovie->year;
cout << "\nYou have entered:\n";
cout << pmovie->title;
cout << " (" << pmovie->year << ")\n";
return 0;
}
|
Enter title: Invasion of the body snatchers
Enter year: 1978
You have entered:
Invasion of the body snatchers (1978)
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The previous code includes an
important introduction: the arrow operator (->).
This is a dereference operator that is used exclusively with pointers to
objects with members. This operator serves to access a member of an object to
which we have a reference. In the example we used:
pmovie->title
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Which is for all purposes
equivalent to:
(*pmovie).title
|
Both expressions pmovie->title and (*pmovie).title are valid and both
mean that we are evaluating the member title of
the data structure pointed by a pointer called pmovie. It must be clearly differentiated from:
*pmovie.title
|
which is equivalent to:
*(pmovie.title)
|
And that would access the
value pointed by a hypothetical pointer member called title. The following panel summarizes possible combinations of
pointers and structure members:
Expression
|
What
is evaluated
|
Equivalent
|
a.b
|
Member
b of object a
|
|
a->b
|
Member
b of object pointed by a
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(*a).b
|
*a.b
|
Value
pointed by member b of object a
|
*(a.b)
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Nesting structures
Structures can also be nested
so that a valid element of a structure can also be on its turn another
structure.
struct movies_t { string title;
int year;
};
struct friends_t { string name;
string email;
movies_t favorite_movie;
} charlie, maria;
friends_t * pfriends = &charlie;
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After the previous
declaration we could use any of the following expressions:
charlie.name
maria.favorite_movie.title
charlie.favorite_movie.year
pfriends->favorite_movie.year
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(where, by the way, the last
two expressions refer to the same member).
Nice
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