Is std::bitset bit-order portable?

Question

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:

• is 0x01 always represented as a byte with bit 7 set?
• is bitset<8>().set(7).to_ulong() always equal to 1?

Answer 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.

Answer 2

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