Re: initializing bitset with variable number of bits

From:
zacariaz@gmail.com
Newsgroups:
comp.lang.c++
Date:
11 May 2007 15:37:27 -0700
Message-ID:
<1178923047.922488.205050@l77g2000hsb.googlegroups.com>
On 12 Maj, 00:23, "Victor Bazarov" <v.Abaza...@comAcast.net> wrote:

zacar...@gmail.com wrote:

The subject isnt very clear, but ill do my best to explain.

In a class in need a bitset which size is defined by the constructor,
e.g. something like this:

class Example {
 std::bitset<?>Bs;


Since the size of the bitset is part of its _type_ definition, bitsets
with different sizes would mean that your 'Example' would have members
of different types. That means those are different 'Example' _types_.
You can only achieve that by making 'Example' a template.

   public:
 Example(int input) {
   ? = input;
 }
};

Now, my knowledge in c++ is simply to limited to work around this.


Actually, C++ is simply limited, not your knowledge. C++ does not
allow to have the same _class_ to contain different types of members
depending on some run-time condition.

I
know that in order to use a variable to initialize a bitset, it has to
be constant, but allso the constant variable has to be initialized
imidiadly, and ofcourse it all has to be done in the right order,
otherwise nothing will make sence or work for that matter. Im confused
and i hope you canhelp.


template<size_t n>
class Example {
    std::bitset<n> Bs;
public:
    ...

};

...
    Example<12> e12;
...

Now, _types_ have to be fixed at the compile time. You cannot create
types during run-time. That's the main limitation.

Perhaps you can roll your own "bitset", instead of using the standard
template... Think 'std::vector<bool>', which has dynamic _size_.


I thought vector<bool> was obsolete, but it might the solution,
however it is my experience, and i might be wrong, that vector<bool>
uses one byte of memory per bool, and if that the case it is
definently not the solution.

Anyhow, thanks for now.

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