[Dev-C++] "const_cast" Help please.....

[Farzan H. <far...@ho...>]

Hi users,

I've recently faced to some confusing results that "const_cast" makes.
Let's assume we have the following code:

const float i=3D55.5;
float * j=3Dconst_cast<float*>(&i);
cout << j << '\t' << &i << endl;  // Displays addresses.
*j=3D45.6;
cout << *j <<'\t' << i << endl;  // Displays contents.

If you run the code you will see something like this:
0x22ff78	0x22ff78
45.6	55.5

As you can see, the address of "i" is "0x22ff78" and it is the place wher=
e =20
"j" points to it.
The value of "i" is 55.5 but the value of the place where "j" points to i=
t =20
is 45.6!!
How can a single memory block store two different values????

Now there is another strange result. If you assign 45.6 to "j" immediatel=
y =20
after
"const_cast" (instead of assigning after displaying addresses) like this:

const float i=3D55.5;
float * j=3Dconst_cast<float*>(&i);
*j=3D45.6;
cout << j << '\t' << &i << endl;  // Displays addresses.
cout << *j <<'\t' << i << endl;  // Displays contents.

the result will be:
0x22ff78	0x22ff78
55.5	55.5

I should mention that this distinction is made when you use DEV-C++ 5.If =
=20
you use DEV-C++ 4,
only the first result will be produced for both codes.

I also tested the codes in MS Visual C++ 6 and it showed me the following=
 =20
result for both:
0012FF7C	0012FF7C
45.6	45.6

The value of "i" is 45.6 now. I think it is more acceptable result becaus=
e =20
by using
"const_cast", a constant storage (i) has been converted to a non constant=
 =20
storage (j)
and you can change its content.

Now please explain these results.
Thanks,



--=20
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