Access List<int> elements and size

[NeoKratox]

Hi,

I wrote this piece of code, basically, it is a vector of lists. But, I can't see why I'm not able do access the elements that I added into list and its size.

Can you help me?

Thanks.

#include "stdafx.h"
#include <list>
#include <vector>
#include <math.h>
#include <cmath>

int _tmain(int argc, _TCHAR* argv[])
{
	int n = 4;
	std::vector<std::list<int>> S;

	int k_max = pow(2,n); 
	S = std::vector<std::list<int>> (k_max);


	for (int k = 1; k <= (k_max-1); k++){
		for(int i = 1; i <= n; i++){
			if (int(floor(k/pow(2,(i-1))))%2 == 1){
				S[k].push_back(i);
				S[k].sort();
			}
		}
	}

	for (int k = 1; k <= (k_max-1); k++){
		for (std::list<int>::iterator it= S[k].begin(); it != --S[k].end(); ++it){
			printf("\n%d",*it);
		}
		//*/
		int tam = S[k].size();
		getchar();
	}


	return 0;
}

[NeoKratox]

Now,

I undestood why it didn't print all itens, the line 27.

it != --S[k].end(); is wrong, ig the list has 2 or less itens, it didn't show them. The right way was:

it != S[k].end();

Now, only the size isn't tottaly accessible.

[abhishekm71]

You are not printing the size().

Also, I can't see any IO header file. I don't know about stdafx.h, does it include IO?

Also, cout is the more C++ way of outputting a simple integer rather than printf().

(sorry for so many Also's)
Also, since you are always sorting the list, IMO you can also use an ordered 'multi-set' rather than a list.

EDIT:
(Another also), there is no need to include <math.h>. It is an old depracted header and <cmath> has all the functionality. You are already including it.

[NeoKratox]

Thanks abhishekm71!

Newbie programmer. I didn't put here the lines where I print, because, they were in portuguese. But the error was that:

printf("item = %d, List size = %d, %d", tam);

And there was no value to be shown...


Thanks!

[LB]

Why are you even using std::list here at all? Considering that you are using it only to store primitive types and you are not utilizing insertion at all, it is a bad choice of container (it is often several orders of magnitude slower than std::vector the way you are using it). std::list has very few specialized use-cases, everywhere else it is the wrong container to use ;)

[NeoKratox]

Really?? Because, I have to create a number of subsets, from a set of integers. So, as I don't know the length of the subset, a list seemed more useful, because for a vector, I would know its size, but, if you have some idea that can help me more, it will be really welcome.

Is better to use cout instead of printf()?

[LB]

Both a list and a vector can grow and shrink dynamically - C++ vector classes are dynamic arrays, they are not related to the mathematical vector.

As for std::cout vs std::printf, it depends on whether you are using C or C++. In C++, you have access to both, but it is a bad idea to mix and match as you please. Pick one kind of I/O - either stick to C I/O, or stick to C++ I/O. As you are working in C++ I highly recommend using C++ I/O (std::cout, std::cin, etc).

[NeoKratox]

Ok, unsderstood. But, how will I do a vector of vectors dynamically?

Here is my code, if you could help me:

        int n, i, u;
	n = 8;
	i = 0;
	u = 4;


	vector<list<int>> S; 

	int k_max = pow(2,n); //máximo de subconjuntos possíveis

	S = vector<list<int>> (k_max);

	for (int k = 1; k <= (k_max-1); k++){ 
		int i_aux = i;
		for(int pos = 1; pos <= n; pos++){
			if (int(floor(k/pow(2,(pos-1))))%2 == 1){
				S[k].push_back(i_aux);
				S[k].sort(); //
				i_aux++;
			}
			else{
				i_aux++;
			}
		}
	}

//Printing
for (int k = 1; k <= (k_max-1); k++){
	cout << "S[" << k << "] = {" ;
	for (list<int>::iterator it = S[k].begin(); it != S[k].end(); ++it){
	      cout << *it << " ";
	}
	cout << "}." << endl;
}

Thanks a lot! But I'm not interested in computational time, for while, I only need this subset structure.