bitset woes

[Paul Richardson]

I have tried reading all the docs and I suppose I am just not parsing
them correctly so I'll ask here

suppose I have three bitsets as follows:
bitset<48> a;
bitset<48> b;
bitset<48> c:

I want to bit-by-bit and the contents of a and b and have them placed
in c, what does the syntax look like for this. I thought, given that
some of the operators are overloads, that i might look like

c = a &b

this does not work however


Thanks in advance

 

[stefan.ciobaca]

It works for me. I'm using g++

g++ (GCC) 3.4.6 (Gentoo 3.4.6-r1, ssp-3.4.5-1.0, pie-8.7.9)

to compile this

<code>
#include <bitset>
#include <iostream>

using namespace std;

int main()
{
bitset<48> a, b, c;

for (int i = 0; i < 48; ++i)
{
b = rand() % 2;
a = rand() % 2;
}
c = a & b;
cout << a << endl << b << endl << c << endl;
}
</code>

What compiler are you using? What exactly doesn't work?

 

[Paul Richardson]

Care to take on the usage of to_string()

It works for me. I'm using g++

g++ (GCC) 3.4.6 (Gentoo 3.4.6-r1, ssp-3.4.5-1.0, pie-8.7.9)

to compile this

<code>
#include <bitset>
#include <iostream>

using namespace std;

int main()
{
bitset<48> a, b, c;

for (int i = 0; i < 48; ++i)
{
b = rand() % 2;
a = rand() % 2;
}
c = a & b;
cout << a << endl << b << endl << c << endl;
}
</code>

What compiler are you using? What exactly doesn't work?

I have tried reading all the docs and I suppose I am just not parsing
them correctly so I'll ask here

suppose I have three bitsets as follows:
bitset<48> a;
bitset<48> b;
bitset<48> c:

I want to bit-by-bit and the contents of a and b and have them placed
in c, what does the syntax look like for this. I thought, given that
some of the operators are overloads, that i might look like

c = a &b

this does not work however


Thanks in advance