use of private members

Here's another clunky experiment that don't work right:

Code:

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#include <iostream>
#include <vector>
using namespace std;

class fibvec  {
        vector<int> series;
        unsigned int len;
        public:
                fibvec() {
                        series.reserve(2);
                        series[0] = series[1] = 1;
                        len = 2;
                }
                int digit (int pos) {
                        int i, j = 0;
                        if (pos >= len) {
                                int nlen = pos+1;
                                series.reserve(nlen);
                                for (i=len;i<nlen;i++)
                                        series[i] = series[i-1] + series[i-2];
                                len = nlen;
                        }
                        return series[pos];
                }
};


int main() {
        int i;
        fibvec eg;

        for (i=5;i<10;i++) cout << eg.digit(i) << '\n';

        return 0;
}

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It doesn't work because within fibvec.digit values for "series" are not kept -- everything is always 0. How can I render this "static" or something? Eg, in perl, a constructor returns a special reference to a hash that contains any number and type of elements (it's special in the sense that the class methods can then be used on it). But I cannot "return" anything with a C++ constructor -- do I have to make fibvec derived from vector (I would assume not, since apparently deriving things from the standard containers is a bad practice).

I have no idea what the problem is you're describing. But one problem is that you'll need resize rather than reserve. Also, why even have a length, isn't that what the .size() function on a vector is for? Of course, that only works if you use resize rather than reserve.

Yeah, you should be using resize() instead of reserve(), and you do not need len. However, suppose you wanted to create a vector of two ints, both set to 1. There is an easy way using an appropriate constructor:

Code:

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std::vector<int> series(2, 1); // 2 ints, both set to 1.

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You can get this same effect for a member variable using the constructor initialisation list:

Code:

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fibvec() : series(2, 1) {}

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Quote:

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Originally Posted by EVOEx

I have no idea what the problem is you're describing. But one problem is that you'll need resize rather than reserve. Also, why even have a length, isn't that what the .size() function on a vector is for? Of course, that only works if you use resize rather than reserve.

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Whew! Using resize instead of reserve made the difference (why?). "This informs the vector of a planned increase in size," -- does that mean you would call reserve then resize? So what is the purpose of reserve?


The len got left in there from an earlier version where I was trying to work around the problem caused by reserve, by using an init method...blah blah blah.

Thanks.

Quote:

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Originally Posted by laserlight

Yeah, you should be using resize() instead of reserve(), and you do not need len. However, suppose you wanted to create a vector of two ints, both set to 1. There is an easy way using an appropriate constructor:

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Okay, but I cannot do that here:

Code:

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class fibvec  {
        vector<int> series(2,1); 

---

for whatever reason.

[edit] alright, I get it: constructor initialisation list

Or just use push_back() instead of resize()?

Okay. Reserve reserves the minimum amount of memory, but doesn't allow the vector to use it yet. Resize actually resizes the vector so that you can use that many elements.

Reserve is useless except for optimization. Let's say you know you'll push_back 1000 times. You could simply call push_back 1000 times and let the vector handle it. However, that might cause the vector to reallocate the memory 1000 times, which is fairly slow. Usually it won't, but it usually will make it reallocate the memory a few times.
To speed up things you can simply call reserve to make sure the vector allocates the required space for those 1000 elements directly. All the subsequent push_backs won't need reallocations (which are quite costly).

So, if you know how many elements are going to be added, and this number is relatively great, it often makes it a bit faster to call reserve first.

Finally, it might be better to make the vector static in the class, as the data is shared over all instances of the class. This would mean that even if you generate thousands of instances of the class, you'll only need one vector. In fact, the class instance might not require any memory at all, then.

Quote:

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Originally Posted by EVOEx

Reserve is useless except for optimization.

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There is another use: to avoid invalidation of iterators, pointers and references due to a reallocation.

Hmmm. This seems to work then:

Code:

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                                int *ref;
                                series.reserve(nlen);
                                for (i=series.size();i<nlen;i++) {
                                        ref = &series.back();
                                        series.push_back(*ref + *(--ref));
                                }

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Altho I would assume it is less optimal than just resizing and using [].

Quote:

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Originally Posted by MK27

Hmmm. This seems to work then:

Code:

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                                int *ref;
                                series.reserve(nlen);
                                for (i=series.size();i<nlen;i++) {
                                        ref = &series.back();
                                        series.push_back(*ref + *(--ref));
                                }

---

Altho I would assume it is less optimal than just resizing and using [].

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I don't think this line is valid, as the behaviour, I think, is undefined:

Code:

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*ref + *(--ref)

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And good point laserlight.

Quote:

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Originally Posted by EVOEx

I don't think this line is valid, as the behaviour, I think, is undefined:

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Yeah, hadn't thought about that. I do get this:
warning: operation on ‘ref’ may be undefined
But thot it was about the range...

I think (and please correct me if I'm wrong, because I'd like to know as well) it is undefined because the c++ standard doesn't specify when the decrement operator happens. Ie
Say ref points to series[3]..

Code:

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series.push_back(*ref + *(--ref));

---

would this mean

Code:

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series.push_back(series[3] + series[2]);

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or

Code:

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series.push_back(series[2] + series[2]);

---

?


As for the implementation of digit(), personally I would do something like

Code:

---

int digit (int pos)
{
        if(pos <= 0) return 0;
        if (pos > series.size())
        {
                if(pos > series.capacity())
                        series.reserve(pos);
                for (int i=series.size();i<pos;i++)
                        series.push_back(series[i-1] + series[i-2]);
        }
        return series[pos-1];
}

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Well, I ended up with something pretty close to that, here's the whole shebang:

Code:

---

class fibvec  {
        vector<int> series;
        public:
                fibvec() : series(2,1) {}
                int digit (unsigned int pos) {
                        if (pos >= series.size()) {
                                int start = series.size(), nlen = pos+1;
                                series.resize(nlen);
                                for (;start<nlen;start++) series[start] = series[start-1] + series[start-2];
                        }
                        return series[pos];
                }
};

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Good job, small bug though ;)
Using Fibonacci number - Wikipedia, the free encyclopedia as reference, your class outputs

Code:

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f(0) = 1
f(1) = 1
f(2) = 2
f(3) = 3
f(4) = 5
f(5) = 8
f(6) = 13
f(7) = 21
f(8) = 34
f(9) = 55
f(10) = 89
f(11) = 144
f(12) = 233
f(13) = 377
f(14) = 610
f(15) = 987
f(16) = 1597
f(17) = 2584
f(18) = 4181
f(19) = 6765
f(20) = 10946

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For any (pos) it outputs the fibonacci number for (pos+1)

Edit: I realized that this might be intentional.. If so just ignore this post :)

Quote:

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Originally Posted by _Mike

For any (pos) it outputs the fibonacci number for (pos+1)

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That's just my ignorance of math -- a set in math starts with an index of 1, not 0, right?

Quote:

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Originally Posted by MK27

That's just my ignorance of math -- a set in math starts with an index of 1, not 0, right?

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It depends on the convention. Your output is perfectly fine. After all, you did not specify exactly what the program was supposed to do, so leaving out 0 is acceptable.