P
Pmb
What does it meant when a function member of a class is declared as const?
Thanks
Pmb
Thanks
Pmb
S
Sharad Kala
Pmb said:
Always check the FAQ first.
http://www.parashift.com/c++-faq-lite/const-correctness.html#faq-18.10
J
John Harrison
Pmb said:
That it does not modify the object on which it is called.
john
K
Karl Heinz Buchegger
Pmb said:
That the function is not going to change the state of the
object when called.
In practice this means: This function can be called on const objects.
P
Pmb
Sharad Kala said:
Thanks. I checked the FAQ and didn't see that. I searched the page for
"const function" and didn't see it.
In any case, after reading it, I don't understand what it means by "The
'abstract (client-visible) state of the object isn't going to change"
Thanks
Pmb
K
Karl Heinz Buchegger
Pmb said:
Exactly what it says.
There is a class.
There is an object of this class.
Now this object is used by some client code.
The client code calls a member function of this object.
When calling the member function, the client code will not
notify any changes in the state of that object.
Example:
I write a class which models a person. The state of that person
consists of its name and its birthdate.
I add a member function to that class which allows you to get
the birthdate. But by calling that function, the object will
not change it's visible state: Neither will the name change
nor will the birthdate change. Thus I will make that function
a const one.
P
Pmb
Karl Heinz Buchegger said:
I don't understand!? Take the program below as an example. The output is
===
Object Test constructed with x = 1, y = 2
x in print() is 8
y in print() is 9
===
The object "test" was declared constant and yet I modified the two data
members x, y. Does that mean that I've changed the value of the object or
the state of the object.
Perhaps I don't understand what is meant above by the "state of the object"?
Thanks
Pmb
--------------------------------------------------------------------------
#include <iostream.h>
class Test{
public:
Test( int = 0, int = 0 );
void print( int , int ) const;
private:
int x;
int y;
};
Test::Test( int i , int j )
{
x = i;
y = j;
cout << "Object Test constructed with x = " << x << ", y = " << y <<
endl;
}
void Test:
rint ( int x, int y ) const{
cout << "x in print() is " << x << endl;
cout << "y in print() is " << y << endl;
}
int main()
{
Test const test( 1, 2 );
test.print ( 8, 9 );
return 0;
}
---------------------------------------------------------------------------
K
Karl Heinz Buchegger
Pmb said:
Where?
You didn't
In the print function you printed the values passed
to that function, not the member variables.
J
John Harrison
Pmb said:
No you didn't. In print x and y are parameters they are not the member
variables x and y. You wouldnot be able to change the member variables x and
y, but you can change the parameters x and y because they are not declared
const.
No you haven't.
object"?
I think you do, the problem it that you don't see that you have two
different x's and two different y's in your program.
john
S
Sharad Kala
Pmb said:
Say you have a class that contains a char pointer as a member. You allocate
memory for that in the constructor. Later in a member function you only change
the contents of the allocated memory . Now as far as client is concerned there
is constness (no change in value of p though what it points to has changed),
hence it can be declared const.
-Sharad
P
Pmb
John Harrison said:
Oops! Thanks
Pmb
S
Simon
Looking at your code bellow
You made the values of x and y equal to 1 and 2.
But not any x and y, the x and y that belongs to the class Test.
To do that you used to values i and j
now your function prints out two values that you have passed 8 and 9 but to
pass the values you called them x and y in your function.
The fact that the class Test also has two place holders called x and y is
irrelevant really, (but bad programming).
If you change your function to
void 'Test:rint ( int i, int j )' const the output will be the value of
the x and y in the class Test, (and i and j would effectively be ignored).
As for a const function it is a function that does not assign values within
it's body.
if you did
void Test:rint ( int i, int j ) const
{
x = j;
cout << "x in print() is " << x << endl;
cout << "y in print() is " << y << endl;
}
it would not work because you are trying to change the value of x. That is
not permitted in a const function.
(note that i oversimplified what happens so you get a better idea of what is
going on).
Simon.
You made the values of x and y equal to 1 and 2.
But not any x and y, the x and y that belongs to the class Test.
To do that you used to values i and j
now your function prints out two values that you have passed 8 and 9 but to
pass the values you called them x and y in your function.
The fact that the class Test also has two place holders called x and y is
irrelevant really, (but bad programming).
If you change your function to
void 'Test:
rint ( int i, int j )' const the output will be the value ofthe x and y in the class Test, (and i and j would effectively be ignored).
As for a const function it is a function that does not assign values within
it's body.
if you did
void Test:
rint ( int i, int j ) const{
x = j;
cout << "x in print() is " << x << endl;
cout << "y in print() is " << y << endl;
}
it would not work because you are trying to change the value of x. That is
not permitted in a const function.
(note that i oversimplified what happens so you get a better idea of what is
going on).
Simon.
J
JKop
Pmb posted:
int DoStuff(const Dog& doggie)
{
doggie.age = 5;
//ERROR, CANNOT EDIT const OBJECT
return 0;
}
-JKop
int DoStuff(const Dog& doggie)
{
doggie.age = 5;
//ERROR, CANNOT EDIT const OBJECT
return 0;
}
-JKop
S
Sharad Kala
JKop said:
His question is that what is a const member function if you read carefully.
J
JKop
Sharad Kala posted:
Opps!!!
int Dog:oStuff(void) const
{
age = 5;
//ERROR, CANNOT EDIT OBJECT, THIS IS A const FUNCTION!!
return 0;
}
Checklist:
1) Does your function edit object member variables? If not, declare it
const .
3) Does your function neither read nor edit member variables? If so, declare
it static .
-JKop
Opps!!!
int Dog:
oStuff(void) const{
age = 5;
//ERROR, CANNOT EDIT OBJECT, THIS IS A const FUNCTION!!
return 0;
}
Checklist:
1) Does your function edit object member variables? If not, declare it
const .
3) Does your function neither read nor edit member variables? If so, declare
it static .
-JKop
A
Andrey Tarasevich
JKop said:
Not correct. For example, take a look at non-const version of
'std::vector:
perator[]'. It doesn't edit any of the object's membervariables. Do you think it should've been declared as 'const'? What
about non-const versions of 'begin()' and 'end()' methods in standard
containers?
Not exactly correct either (for similar reasons).
BTW, where is 2) ?