bitset assignment

[Andy Skypeck]

I have a question concerning the use of the [] operator in bitset
asignment.

bitset<4> a("1100");
bitset<4> b("0011");

a[i] = a[j];

Is clearly supported by the standard, and works with all the C++
compilers I have worked with. But what about

a[i] = b[j]; // ?

Assume that there are no indices out of range.
This works with gcc-3.* and MSVC 6 and higher but fails on an older
Borland compiler. I contend that this is ok because both a[i] and
b[j] return type bitset<N>::reference. Others say no, but for
unspecified reasons. Does anybody
have an authoritative answer?

[Ron Natalie]

"Andy Skypeck" <as******@earthlink.net> wrote in message news:42**************************@posting.google.c om...

Is clearly supported by the standard, and works with all the C++
compilers I have worked with. But what about

a[i] = b[j]; // ?

Assume that there are no indices out of range.

It's supposed to work. That's the whole point of operator[] returning bitset::reference. Otherwise
it'd just return bool. Bitset::reference has a copy assignment operator.

[Dan W.]

On 3 Dec 2003 08:45:34 -0800, as******@earthlink.net (Andy Skypeck)
wrote:

I have a question concerning the use of the [] operator in bitset
asignment.

bitset<4> a("1100");
bitset<4> b("0011");

a[i] = a[j];

Is clearly supported by the standard, and works with all the C++
compilers I have worked with. But what about

a[i] = b[j]; // ?

Assume that there are no indices out of range.
This works with gcc-3.* and MSVC 6 and higher but fails on an older
Borland compiler. I contend that this is ok because both a[i] and
b[j] return type bitset<N>::reference. Others say no, but for
unspecified reasons. Does anybody
have an authoritative answer?

Try casting b[1], say, to an int, and see what the different compilers
give you; assuming it's set, it should return 1. If it returns 2 it
probably means they took a shortcut that's confusing the assignment,
--i.e.: broken compiler.