I'd like to use the shift operators as doing a bit rotation instead of their actual bit shift. This is my expected behavior:
std::bitset<8> b8("1010"); // b8 = 00001010
b8 <<= 5; // b8 = 01000001
So I try and overloaded the <<= operator, referencing the bitset definition, as followed:
#include <iostream>
#include <bitset>
using namespace std;
template <size_t size>
bitset<size>& bitset<size>::operator<< (size_t pos) noexcept { // an error at here
}
I've got an error at keyword operator:
Out-of-line definition of 'operator<<' does not match any declaration in 'bitset<_Size>'
How can I fix it? My env. is:
llvm-g++ -v) : Apple LLVM version 9.0.0 (clang-900.0.38)std::bitset::operator<<= is a member function of the template class std::bitset. You cannot redefine this operator. And you cannot even hide it with another:
template <std::size_t size>
std::bitset<size>& operator<<=(std::bitset<size>& bitset, std::size_t count) noexcept {
// ...
return bitset;
}
This compiles but achieves nothing since when you write b8 <<= 5, unqualified-id lookup finds std::bitset::operator<<= before considering your free function.
You should use another operator, define a rotate function, or define a wrapper rotate class:
struct rotate
{
rotate(std::size_t n) : value(n) {}
std::size_t value;
};
template <std::size_t size>
std::bitset<size>& operator<<=(std::bitset<size>& bitset, rotate r) noexcept {
bitset = bitset << r.value | bitset >> (size-r.value);
return bitset;
}
usage:
std::bitset<8> b8("1010"); // b8 = 00001010
b8 <<= rotate(5); // b8 = 01000001