Ce cours initie aux bases de la programmation en utilisant le langage C++ : variables, boucles, fonctions, ... Il ne présuppose pas de connaissance préalable. Les aspects plus avancés (programmation orientée objet) sont donnés dans un cours suivant, «Introduction à la programmation orientée objet (en C++)». Il s'appuie sur de nombreux éléments pédagogiques : vidéos sous-titrées, quizz dans et hors vidéos, exercices, devoirs notés automatiquement, notes de cours.
Puissance 4 : fonctions est_ce_gagne et compte
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Computer Programming, C++11, C++, Programming Language
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LL
Dec 24, 2016
J'ai commencé ce cours pour repartir sur de bonne bases en programmation c++ et je dois dire que je l'ai trouvé très clair et très complet.
MM
May 17, 2018
Très bon enseignement. Je suis pas pleinement à la hauteur du dernier devoir mais j'ai bien avancé tout de même. Merci
De la lección
Etude de cas
Nous voulons terminer notre cours avec une étude de cas, la création d'un jeu de « Puissance 4 », nous permettant de revoir tous les concepts abordés au long du cours. Nous en profitons pour présenter un dernier type de données : le type « énuméré ».
Impartido por:
Jean-Cédric Chappelier
Dr.
Jamila Sam
Dr
In the previous episode, we have seen how to write the main loop,
allowing the players to take their turns and play.
Only missing
was the stopping condition.
Thus, we still need to write
something here so that the loop will stop
either once a player has won or once the grid is full.
For now, we will focus on
the case where we expect the loop to stop
because one of the two players has won.
To that end, we will simply introduce
a new function called "est_ce_gagne" (TN: means "is_it_won").
We will call this functions as follows.
We will store the result of the "est_ce_gagne" function
inside a variable called "gagne" (TN: means "won" or "wins").
This bool-type variable
"gagne" will receive
the return value of the function "est_ce_gagne".
As argument, the "est_ce_gagne" function
will receive the grid,
and the color of the player
who has just played.
Also, we will need to declare our "gagne" variable.
Since we will need to use it
as the stopping condition, we need to declare "gagne" outiside the loop,
that is, somewhere around here. Now, we can use
the "gagne" variable in our stopping condition.
This will simply be
written like this.
We will repeat the loop
as long as the player has not won.
We now need to write this "est_ce_gagne" function; this will probably
be the trickiest part of our program.
So, how should
we proceed?
Actually, there are several ways to write this "est_ce_gagne" function.
The strategy we are offering is the following:
We will run through the grid row by row
and column by column
until we encounter a disk which color
is the color of the player who has just played.
Let us say, for example, that the red player has just played.
This newly-dropped disk is here.
Now, we will start from this disk
and look over the grid in several directions,
horizontally, vertically and diagonally.
Thus, we will count the
same-colored disks,
starting from this one here.
If, for example, we consider this direction here,
we can count
1, 2, 3, 4 red disks;
the red player has thus won.
If, on the other hand, we have looked over the whole grid
without counting at least 4 disks,
we know that the player has not won.
By the way, we do not have to consider
all the directions.
Actually, we will only have to count the disks in 4 directions.
For example, we will choose
the Northeast,
the East,
the Southeast,
and the South.
But why is
it possible to consider
these 4 directions only?
We said that we were going to look over the grid
line by line
and column by column.
Let' assume once more that the red player has played,
the first disk we have found
is this one.
We will now start counting the disks in this direction here;
we have only counted one disk, this one.
Same thing for this direction here and this direction here;
there is only one disk in these 3 directions.
In this direction here, we count 2 disks, which is still not enough to win.
We will now keep looking over the grid.
The next red disk
we will find
is this one, in this direction here.
In this direction, we will only count
2 disks, which is not enough to win. In this direction
and this direction here, we will only count one disk:
still not enough to win.
Thus, we will keep looking over the grid
until we reach this disk here.
We will count the disks
in this direction here.
This time, we will count 1, 2, 3, 4 disks;
we thus know that the red player has won.
In order to prove that the red player has indeed aligned 4 disks,
we can either start from this disk here, and go in this direction here
or start from this disk here and go in this direction here.
Therefore, we only need to consider
one of the two directions. And the same goes
for the 6 other directions.
We will only consider this direction here
and not this direction here.
We will consider this direction here
and not this one.
This direction here and not this one.
Therefore, we only need to consider 4 directions.
Now, we will need to code this strategy and write the "est_ce_gagne" function.
Be reminded that we will call the "est_ce_gagne" function
as follows.
We will store
the return value of the function inside a bool-type variable.
Also, the function receives the grid
and the color of the player who has just played
as its two parameters.
The header of this function will thus be this one here.
The result will be a bool-type value.
We will pass the grid
by constant reference since the grid does not need to be modified.
In this "est_ce_gagne" function,
the second parameter is a "Couleur"-type variable.
Our strategy is to look over the grid.
We will do just do that
thanks to two "for loops".
Since we need the disk coordinates,
we will use two "old-style for loops"; these will be written as follows.
The variables "ligne" et "colonne" (TN: means "row" and "column)
will contain the coordinates of the disk from which
we will start counting the aligned disks. In order to simplify
the writing of this code, we will introduce
a local variable
called "couleur_case" (TN: means "square_color").
This variable will contain the color of the disk.
If the square color is the same
as the color of the player who has just played,
we will count the disks aligned, starting from this square,
in the 4 directions
we are focusing on.
Since counting the aligned disks is no easy task,
we will introduce a new function called "compte" (TN: means "count).
What arguments does this "compte" function need?
First of all, of course, the grid.
Also needed is the square from which we will count the aligned disks;
this corresponds simply
to the variables "ligne" and "colonne".
We will also need the direction in which
we are counting the disks.
So, how can we define
this direction?
Let us suppose that we wish to count the disks
in the Northeast direction.
Then, in order to move from one square to the next,
we will subtract 1 from the row
and add 1 to the column.
Thus, we will use these variables here: this -1 and this +1
as arguments of the "compte" function.
This means that we will be counting in the Northeast direction.
The "compte" function will return the
number of same-colored
aligned disks.
We know that if this number of aligned disks
is greater or equal to 4,
the player has won.
But why greater or equal to 4
instead of simply 4? Of course because
we may and can align more than 4 disks.
Obviously, we will also need to consider the 3 other directions.
We will thus proceed this way.
In order to count the disks horizontally,
we will not change the "ligne" here, thus using 0
as argument here.
On the other hand,
we will move onto the next column
which is why we use the value +1 as the last argument.
Now to
count the disks
in the Southweast
direction
In order to move from one square to the next,
we will add 1 to the row
-- thus using the value +1 here--
and we will also add 1 to the column
-- thus using the value -1 here.
We still need to deal with the very last direction.
In order to count disks in the South direction,
we will move from one square to the next
by adding 1 to the row
-- thus using the value +1 here --
but the column will not change -- thus using the value 0 here.
If one of these 4 conditions is true,
we know that the player has won.
Since only one of these conditions needs to be true,
we used "or" here.
In this case, our function will return true.
We here close the previously opened braces.
Reaching this point of our function,
this means that we have looked over the whole grid
without ever passing by here.
Indeed, if we had, we would already have exited the function.
We therefore know
that we were not able to count 4
or more aligned disks; thus, the player has not won, yet.
The function will thus return false.
Obviously, we still have to write the "compte" function.
We are calling this "compte" function here, for example.
It will return the number of aligned disks
which is for sure a non-negative integer,
which is why we used the "unsigned int" type in order to define the return type of the function.
The first parameter of the function is thus the grid,
passed, once again,
by constant reference.
The two following parameters are simply
the row and the column
of the square from which we will start counting the disks.
The last two parameters
of the function
are the two values which will define the couting direction.
These can be positive or negative integers; we will thus use the int type.
We will call these parameters "dir_ligne"
and "dir_colonne" in order to signify that these two values
will correspond to the row and column
for a given direction.
We will resort to a variable "compteur" (TN: means "counter")
initialized to 0. Its type is the same as the return type of the "compte" function.
It will keep the count of aligned disks.
What should the "compte" function do exactly?
It must start from the square with the coordinates "ligne_depart"
and "colonne_depart". (TN: means "starting_row" and "starting_column")
It must loop over the grid following a given direction
as long as it encounters disks which color is the same as
the disk located on "ligne_depart"
and "colonne_depart".
How will we do this?
We will use variables "ligne" and "colonne" (TN: means "row" and "column")
which will be initialized to
"ligne_depart"
and "colonne_depart".
At every loop,
in order to move from one square to the next,
we will add "dir_ligne" to the variable "ligne"
and add "dir_colonne" to the variable "colonne".
Since we will look over the grid
as long as we encounter disks which color is the same as the disk
in "ligne_depart"
and "colonne_depart"
we will use a "while loop".
At every loop,
we will have found a new disk of the selected color
and will thus increment the "compteur" variable.
How will we writte all this? Well, we will start by declaring
the variables "ligne" and colonne"
initialized to "ligne_depart"
and "colonne_depart".
As we said, we are using a "while loop" with a stopping condition
depending on the color of the disk on "ligne" and "colonne".
We will keep going as long as the color of this disk is the same as
the color of the disk on "ligne_depart"
and "colonne_depart".
At every loop we will thus increment the "compteur" variable
since we have found
a new disk which color is the same as the first one.
Then, we will add
"dir_ligne"
and "dir_colonne"
to "ligne" and "colonne".
By the way, please note that we haved added blank spaces here
and here in order to align the " = " and " + " signs on these two lines;
this makes the code easier to read
and makes it obvious that
the variables "ligne" and "colonne" share the same nature.
When we exit the loop, this means that we have found a disk
which color was not the good one.
We can thus return the value contained in the "compteur" variable.
However, this is not entierely correct.
Do you see why?
At this point,
similarly to the "joue" function,
if we do not add conditions to our "while loop",
the possibility remains that we could exit the "grille" array.
We will follow the path opened with the alternative version of the "joue" function
and use the same solution.
Namely, we will add a test regarding the variable "ligne",
making sure that it remains less than the size of "grille".
The same goes for the variable "colonne".
Finally, we will return to our "est_ce_gagne" function and notive that,
if the row of the starting square
has 0, 1 or 2 as index,
there is not need to search
in this direction here, for example.
Indeed, we already know
that the player could not possiblity align four disks in this direction here.
We could align 3 disks at most,
if we are starting from the row number 2.
The same goes if we are starting
from the edge of the grid
and are counting the disks towards the right
or if we are starting from the bottom of the grid
and counting the disks downward.
Therefore, the "est_ce_gagne" function executes more operations than necessary.
We will thus add conditions preventing the disk counting
whenever we are sure 4 disks could not possibly be aligned.
This will be written as follows:
We are considering the Northeast direction.
So that at least 4 disks could be aligned,
the row needs to be
strictly greater than 2.
Also, the column should not be too big;
feel free to check that, in fact,
the column must be less than
the number of columns minus 4.
This will be written
as follows.
If we are searching for disks aligned horizontally,
we only nest to test the column.
The test will be the same as before.
The test for the
Southeast direction
will be written like this.
This time, we will make sure that that neither the row
nor the column are too big.
In the case where we count the disks in the South direction,
we simply need to verify that the row is not too big.
Now, as you can see, we have introduced several repetitions.
Indeed, this expression appears here, here and here.
This expression also appears here.
In order to avoid this, we will introduce two constants
which will contain the values of these expressions.
This will be written like this.
We will call our constants "ligne_max" (TN: means "max_row")
and "colonne_max" (TN: means "max_column").
We will use them
wherever the previous expressions were used.
Voilà! This is now the final version of our
"est_ce_gagne" and "compte" functions.
Now, we only need to deal with the situation where the grid becomes full
without any player having won.
This will be the subject of the very last video.
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