Using STL how do I convert a variable to a binary string

[young_leaf]

Using STL how do I convert a variable to a binary string but with
special case which
for example BYTE x, i need only the first 3 bits of this byte, is that
possible?

to be more specific,
from that byte i want to generate (so valid values for x will be 0-5)

000
001
010
011
100
101
110 - not used
111 - not used

--
yl

[JustBoo]

On 24 Jan 2006 10:48:39 -0800, "young_leaf" <ke***********@gmail.com>
wrote:

Using STL how do I convert a variable to a binary string but with
special case which
for example BYTE x, i need only the first 3 bits of this byte, is that
possible?

std::bitset

I quite like it. You can manipulate each bit in a simple way.

HTH - Good Luck

[Thomas Tutone]

young_leaf wrote:

Using STL how do I convert a variable to a binary string but with
special case which
for example BYTE x, i need only the first 3 bits of this byte, is that
possible?

to be more specific,
from that byte i want to generate (so valid values for x will be 0-5)

000
001
010
011
100
101
110 - not used
111 - not used

If by "STL," you mean the C++ standard library, then I think a
std::bitset does what you need, and specifically its to_string() member
function. You'll need to mask or reset the bits you don't need.

Best regards,

Tom

[Victor Bazarov]

young_leaf wrote:

Using STL how do I convert a variable to a binary string but with
special case which
for example BYTE x, i need only the first 3 bits of this byte, is that
possible?
Everything is possible.

Use 'ostringstream'. Write two functions: one to convert 'BYTE', and the
other to convert everything else. The one that converts 'BYTE' should
also check the range.
to be more specific, [...]

Yeah, yeah, we got that.

V

[Mike Wahler]

"young_leaf" <ke***********@gmail.com> wrote in message
news:11**********************@o13g2000cwo.googlegr oups.com...

Using STL how do I convert a variable to a binary string but with
special case which
for example BYTE x, i need only the first 3 bits of this byte, is that
possible?

to be more specific,
from that byte i want to generate (so valid values for x will be 0-5)

000
001
010
011
100
101
110 - not used
111 - not used

#include <bitset>
#include <climits>
#include <iostream>
#include <string>

unsigned int bin_digits(unsigned int i)
{
unsigned int result(1); /* count 0 as 1 bit */

while(i--)
{
i /= 2;
++result;
}

return result;
}

std::string gen(unsigned int x, unsigned int mx = 5)
{
std::bitset<sizeof x * CHAR_BIT> bs(x);
std::string s(bs.to_string());

return (x <= mx) ? s.substr(s.size() - bin_digits(mx))
: "out of range";
}

int main()
{
for(int i = 0; i < 10; ++i)
std::cout << i << " : " << gen(i) << '\n';

return 0;
}

Output:

0 : 000
1 : 001
2 : 010
3 : 011
4 : 100
5 : 101
6 : out of range
7 : out of range
8 : out of range
9 : out of range
-Mike

[Pete Becker]

Mike Wahler wrote:

#include <bitset>
#include <climits>
#include <iostream>
#include <string>

unsigned int bin_digits(unsigned int i)
{
unsigned int result(1); /* count 0 as 1 bit */

while(i--)
{
i /= 2;
++result;
}

return result;
}

std::string gen(unsigned int x, unsigned int mx = 5)
{
std::bitset<sizeof x * CHAR_BIT> bs(x);
std::string s(bs.to_string());

return (x <= mx) ? s.substr(s.size() - bin_digits(mx))
: "out of range";
}

int main()
{
for(int i = 0; i < 10; ++i)
std::cout << i << " : " << gen(i) << '\n';

return 0;
}

Seems awfully complicated for such a simple transformation.

#include <string>
#include <iostream>

std::string to_binary(unsigned value)
{
if (5 < value)
return "out of range";
char res[4];
res[3] = '\0';
res[2] = (value & 0x01) ? '1' : '0';
res[1] = (value & 0x02) ? '1' : '0';
res[0] = (value & 0x04) ? '1' : '0';
return res;
}

int main()
{
for(int i = 0; i < 10; ++i)
std::cout << i << " : " << to_binary(i) << '\n';

return 0;
}

Here's one that's even simpler:

char *binary[7] =
{ "000", "001", "010", "011", "100", "101", "out of range" };

std::string to_binary(unsigned value)
{
if (6 < value)
value = 6;
return binary[value];
}

Of course, neither of these honors the artificial constraint that the
code should use STL, if "STL" means "Standard Template Library".

--

Pete Becker
Dinkumware, Ltd. (http://www.dinkumware.com)

[roberts.noah]

Pete Becker wrote:

Of course, neither of these honors the artificial constraint that the
code should use STL, if "STL" means "Standard Template Library".

You could always do something like this:

std::vector<int> v;
int b = 42;
do
{
v.push_back(i & 1 ? 1:0);
}
while (b >> 1)

then...
std::ostringstream out;
std::copy(v.begin(), v.end(), ostream_iterator(out));

std::string result = out.str();

Yeah, it might be about the most inefficient way you could possibly do
it (ok, I could probably come up with worst) but at least it uses the
STL. ;)