Does C++ say anything on bit-ordering? I'm especially working on protocol packet layouts, and I'm doubting whether there is a portable way to specify that a certain number be written into bits 5,6,7, where bit 5 is the 'most significant'.
My questions:
bitset<8>().set(7).to_ulong() always equal to 1?
From 20.5/3 (ISO/IEC 14882:2011)
When converting between an object of class bitset and a value of some integral type, bit position pos corresponds to the bit value 1 << pos.
That is, bitset<8>().set(7).to_ulong() is guaranteed to be (1 << 7) == 128.
bitset doesn't do serialization, so you don't (need to) know. Use serialization/deserialization.
is bitset<8>().set(7).to_ulong() always equal to 1
No, not on my machine (see below).
However, I'd certainly expect the iostream operators to behave portably:
#include <bitset>
#include <sstream>
#include <iostream>
int main()
{
std::bitset<8> bits;
std::cout << bits.set(7).to_ulong() << std::endl;
std::stringstream ss;
ss << bits;
std::cout << ss.rdbuf() << std::endl;
std::bitset<8> cloned;
ss >> cloned;
std::cout << cloned.set(7).to_ulong() << std::endl;
std::cout << cloned << std::endl;
}
Prints
128
10000000
128
10000000
If the question is whether you can happily ignore endianness of the platform while sending binary objects over the network, the answer is you cannot. If the question is whether the same code compiled in two different platforms will yield the same results, then the answer is yes.
