[RESOLVED] An error in my code - binary digit matrix implementation by using 2D vector

[Kmilano]

Hello

[/HTML]

[VictorN]

...
when I use vectors which I commented in the following code without void function everything is okay and if I want to merge them into one vector like following with void function I will have an error in this line:

for (const auto& row : generate_mtx(nbits))

1. Please, post the exact error text.
2. Please, use CODE tags, not the HTML ones!

[Kmilano]

the errors:
1
main.cpp||In function ‘int main()’:|
2
main.cpp|40|error: invalid initialization of reference of type ‘std::vector<std::vector<int> >&’ from exp<b></b>ression of type ‘size_t {aka long unsigned int}’|
3
main.cpp|12|note: in passing argument 1 of ‘void generate_mtx(std::vector<std::vector<int> >&, std::size_t)’|
4
main.cpp|42|error: unable to deduce ‘auto&&’ from ‘row’|
5
||=== Build failed: 2 error(s), 0 warning(s) (0 minute(s), 0 second(s)) ===|

[VictorN]

the errors:
1
main.cpp||In function ‘int main()’:|
2
main.cpp|40|error: invalid initialization of reference of type ‘std::vector<std::vector<int> >&’ from exp<b></b>ression of type ‘size_t {aka long unsigned int}’|
...

...
		for (const auto& row : generate_mtx(nbits))
		{
			for (int v : row) std::cout << v << ' ';
			std::cout << '\n';
		}
...

Your code refers to the function generate_mtx(nbits), however the only function with this name is declared as void, so it cannot return anything...
Sure, it is not what you expected!

[Kmilano]

in fact,

[VictorN]

in fact, first my code was like the following code and it works correctly, but because I must use void function to match base class with a derived class, so I changes that line to void function,
now, what's a solution to solve this error?

You could implement the overloaded method in your derived class that would either return your created vector or you would pass this vector as a parameter by reference into your method.
Variant 1 (pseudo code):

typedef std::vector< std::vector<int> > myMatrix;

myMatrix generate_mtx(std::size_t nbits)
{
	nbits %= std::numeric_limits<unsigned long long>::digits;

	myMatrix result(nbits);

	// note: col with all zeroes is skipped (start with number = 0 to include it)
	for (unsigned long long number = 1; number < ubound(nbits); ++number)
	{
		auto n = number;
		for (auto& vec : result)
		{
			vec.push_back(n % 2);
			n /= 2;
		}
	}
	std::reverse(std::begin(result), std::end(result));

	return result;
}

Variant 2 (pseudo code):

typedef std::vector< std::vector<int> > myMatrix;

myMatrix generatedMatrix;
void generate_mtx(std::size_t nbits, myMatrix& generatedMatrix)
{
	nbits %= std::numeric_limits<unsigned long long>::digits;

	// note: col with all zeroes is skipped (start with number = 0 to include it)
	for (unsigned long long number = 1; number < ubound(nbits); ++number)
	{
		auto n = number;
		for (auto& vec : generatedMatrix)
		{
			vec.push_back(n % 2);
			n /= 2;
		}
	}
	std::reverse(std::begin(generatedMatrix), std::end(generatedMatrix));
}

[2kaud]

for (const auto& row : generate_mtx(nbits))

When using a range-based for loop, the range expression (the return type of generate_mtx(nbits) in this case) needs to be of a type that supports the begin() and end() iterator functions so that the range can be obtained. vector<> does but void doesn't. If generate_mtx() is required to return a void, then generate_mtx() needs to have a reference parameter as Victor uses in Variant 2 in post #6. In this case the range_for would be something like (not tried)

typedef std::vector< std::vector<int> > myMatrix;

myMatrix generatedMatrix;
generate_mtx(nbits, generatedMatrix);

for (const auto& row : generatedMatrix)
{
    for (int v : row)
        std::cout << v << ' ';

    std::cout << '\n';
}

[Kmilano]

thank you for your response, but I cannot use matrix and I must use the code which exists in my first post, In my last reply I posted my original correct code, but I cannot use it,
if its possible please give me an alternative way for correcting the first code:

[VictorN]

[QUOTE=Kmilano;2221795]thank you for your response, but I cannot use matrix
What "matrix" do you mean?
if it is the one I referred in my post as

typedef std::vector< std::vector<int> > myMatrix;

then sorry! It is just a typedef! I used it to make code simpler (to write and to understand!)

if its possible please give me an alternative way for correcting the first code:

#include "stdafx.h"
#include <iostream>
#include <limits>
#include <vector>
#include <algorithm>
using namespace std;

constexpr unsigned long long ubound(std::size_t nbits)
{
	if (nbits < 2) return 2;
	else return ubound(nbits - 1) * 2;
}

//std::vector< std::vector<int> > generate_mtx(std::size_t nbits)
void generate_mtx(std::vector< std::vector<int> >& result, std::size_t nbits)
{
	nbits %= std::numeric_limits<unsigned long long>::digits;

	//std::vector< std::vector<int> > result(nbits);

	// note: col with all zeroes is skipped (start with number = 0 to include it)
	for (unsigned long long number = 1; number < ubound(nbits); ++number)
	{
		auto n = number;
		for (auto& vec : result)
		{
			vec.push_back(n % 2);
			n /= 2;
		}
	}

	// to get the rows in the same order as illustrated in the example  
	std::reverse(std::begin(result), std::end(result));

	//return result;
}

int main()
{

	if (size_t nbits = 4)
	{
		std::cout << "check matrix (H) is: " << std::endl;

		for (const auto& row : generate_mtx(nbits))
		{
			for (int v : row) std::cout << v << ' ';
			std::cout << '\n';
		}


	}
	int pause;
	std::cin >> pause;
	//return 0;
}

I did it in my previous post.

Alternatively you could define your "matrix" (std::vector< std::vector<int> >) as a global object but in C++ it is not recommended.

[2kaud]

I think this is what you are after. It produces the required output.

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

constexpr unsigned long long ubound(std::size_t nbits)
{
	if (nbits < 2) return 2;
	else return ubound(nbits - 1) * 2;
}

void generate_mtx(std::vector< std::vector<int> >& result, std::size_t nbits)
{
	nbits %= std::numeric_limits<unsigned long long>::digits;
	result.resize(nbits);

	// note: col with all zeroes is skipped (start with number = 0 to include it)
	for (unsigned long long number = 1; number < ubound(nbits); ++number)
	{
		auto n = number;
		for (auto& vec : result)
		{
			vec.push_back(n % 2);
			n /= 2;
		}
	}

	// to get the rows in the same order as illustrated in the example
	std::reverse(std::begin(result), std::end(result));
}

int main()
{
	if (size_t nbits = 4)
	{
		std::vector< std::vector<int> > results;

		std::cout << "check matrix (H) is: " << std::endl;
		generate_mtx(results, nbits);

		for (const auto& row : results)
		{
			for (int v : row) std::cout << v << ' ';
			std::cout << '\n';
		}
	}

	std::cin.get();
}

check matrix (H) is:
0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
0 0 0 1 1 1 1 0 0 0 0 1 1 1 1
0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

[Kmilano]

yes thank you so much, its my desired code