bitset

[Anders]

Hello.

Can anyone tell me why you have to explicitly write the template parameters
in the .to_string() method? e.g:

bitset<8> bit(string("11"));
cout << bit.to_string<char, char_traits<char>, allocator<char> >() << endl;

...

 

[John Harrison]

Hello.

Can anyone tell me why you have to explicitly write the template parameters
in the .to_string() method? e.g:

bitset<8> bit(string("11"));
cout << bit.to_string<char, char_traits<char>, allocator<char> >() << endl;

Noramlly which version of a template function is called is worked out based
on the types of the parameters of that function. Since to_string does not
have any parameters you have to tell the compiler explicitly which kind of
string you want returned.

john

 

[Victor Bazarov]

Can anyone tell me why you have to explicitly write the template parameters
in the .to_string() method? e.g:

bitset<8> bit(string("11"));
cout << bit.to_string<char, char_traits<char>, allocator<char> >() <<

endl;

Because 'to_string' is a template member. You only need to specify
the first one, by the way: bit.to_string<char>()

However, *I* don't have to do that to output the actual bitset:

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

using namespace std;

int main()
{
bitset<8> bit(string("11"));
cout << bit << endl;
}

works fine for me. It prints
00000011

Victor

 

[Ron Natalie]

Hello.

Can anyone tell me why you have to explicitly write the template parameters
in the .to_string() method? e.g:

bitset<8> bit(string("11"));
cout << bit.to_string<char, char_traits<char>, allocator<char> >() << endl;

Functions are not overloaded by return value. It doesn't know what kind of
string you wanted to convert it to.