Why a C++ bitset initialized using an string is reversed?

Question

For the first time, I initialized a bit set using a string and found out that the bits are stored in reverse order, i.e.:

bitset<3> test(string("001"));

then the bits are stored as bellow: test[0] = 1 test[1] = 0 test[2] = 0

I am not sure if I'm doing something wrong or this is the way it should be.

That's how bits are usually numbered: the bit on the right side is bit zero, because its value is 2^0. The second bit from the right is bit one because its value is 2^1. And so on. — R. Martinho Fernandes, Jun 27 '13 at 12:29
In other words, it isn't reversed. The 0th bit is set to 1, the rest to 0, and the indexing respects that convention. — juanchopanza, Jun 27 '13 at 12:44
Voted to close as "unclear what you are asking". You ask why something is that isn't. — juanchopanza, Jun 27 '13 at 12:58
@juanchopanza: Then I don't understand why in your first comment you would answer a question which you think is unclear? — dahma, Jun 27 '13 at 13:29
It is not unclear. It is a non-question: "**Why a C++ bitset initialized using an string is reversed?**". It is not reversed, so there is no why. — juanchopanza, Jun 27 '13 at 13:30
If I initialize something with "001" and get "100" when I print it, for example, I would say it is in reverse order. — dahma, Jun 27 '13 at 13:35

score 3 · Accepted Answer · answered Jun 27 '13 at 12:57

This is the way it should be. Bits stored in a bitset are ordered in such a way so that the index of a bit is the factor it is raised by.

In other words, the value at test[0] is the 2^0 bit, test[1] is 2^1, test[2] is 2^2, etc.

Endianness has nothing to do with it.

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