std::bitset construction with std::string question

[Dill Hole]

Can anyone tell me why

std::bitset<2> foo(std::string ("01"))

initializes the bitset in reverse order, i.e.

foo[0]=true
foo[1]=false

I would expect the bitset to be initialized in string text order.

[John Harrison]

Re: std::bitset construction with std::string question


"Dill Hole" <dillhole@micro soft.com> wrote in message
news:7brbgvs98i bjn2fcdtl7drc7t k82fstpoj@4ax.c om...[color=blue]
> Can anyone tell me why
>
> std::bitset<2> foo(std::string ("01"))
>
> initializes the bitset in reverse order, i.e.
>
> foo[0]=true
> foo[1]=false
>
> I would expect the bitset to be initialized in string text order.[/color]

I wouldn't,

6 = 110 = 1*2^2 + 1*2^1 + 0*2^0 == foo[2] is 1, foo[1] is 1, foo[0] is 0.

In other words its a consequence of the fact that in English numbers the
most signficant digit goes first.

john

[Mike Wahler]

Re: std::bitset construction with std::string question


Dill Hole <dillhole@micro soft.com> wrote in message
news:7brbgvs98i bjn2fcdtl7drc7t k82fstpoj@4ax.c om...[color=blue]
> Can anyone tell me why
>
> std::bitset<2> foo(std::string ("01"))
>
> initializes the bitset in reverse order, i.e.
>
> foo[0]=true
> foo[1]=false
>
> I would expect the bitset to be initialized in string text order.[/color]

See John H's reply, then observe the output of:

cout << foo << '\n';

-Mike