bitset variable

[rudra]

Dear friends,
i am in mess with bitset again. i have managed to write a messy code
which checks the machine bitsize and then wants to do a bitflip. here
is the code:
int bit;
if (strncmp(sysptr, "i686",4)==0)
{
bit=32;
}
......
bitset<32i (0);
....etc.
where in bitset is equal to bit,which is 32. but i tried to write the
bitset as *bitset<biti (0);*(i.e. to read the bit)
where its giving the error:error: ‘bit’ cannot appear in a constant-
expression
can you tell me a way to make that possible?

[Ian Collins]

rudra wrote:

Dear friends,
i am in mess with bitset again. i have managed to write a messy code
which checks the machine bitsize and then wants to do a bitflip. here
is the code:
int bit;
if (strncmp(sysptr, "i686",4)==0)
{
bit=32;
}
......
bitset<32i (0);
....etc.
where in bitset is equal to bit,which is 32. but i tried to write the
bitset as *bitset<biti (0);*(i.e. to read the bit)
where its giving the error:error: ‘bit’ cannot appear in a constant-
expression
can you tell me a way to make that possible?

You can't. Template (integer) arguments must be compile time constants.

--
Ian Collins.

[Jerry Coffin]

In article <1efb223b-6b78-40f7-b4f7-8f7a34c1adb2
@w7g2000hsa.googlegroups.com>, bn********@gmail.com says...

[ ... create a bitset where the size isn't a compile-time constant ]

can you tell me a way to make that possible?

Boost Dynamic Bitset.

--
Later,
Jerry.

The universe is a figment of its own imagination.