array duplications

[cdg]

Could anyone tell me how to write a loop for an array, that would have to
check each iteration for duplications of previous entered values. So, the
exact number of loops is not known, but the total number of array elments is
known.

Example: array size - array[100]

(some type of loop statement)
{
array[n] "equal to" or "not equal to" a variable. If not a duplication,
increase the count by one. If it is a duplication, check the next variable
for duplication and so on.
}

This is only approach I have tried, but it is not working.

while(i<100)
{
ct += 1;

"statements "

for(n=0; n<ct; n++)
{
array[n] != var.name ? array[n] == var.name : n--
}
}

[Daniel T.]

In article <sl************ **********@FIAD 06.norwich.edu> ,
Neil Cerutti <le*******@emai l.com> wrote:

On 2006-02-24, Daniel T. <po********@ear thlink.net> wrote:

In article <sl************ **********@FIAD 06.norwich.edu> ,
Neil Cerutti <le*******@emai l.com> wrote:

On 2006-02-24, cdg <an****@anywher e.com> wrote:
> I`ll try to describe exactly what is needed for the algorithm.

> This is a function that receives "unsigned integers". Then these go
> thru some processing steps. And result in "integers" within a
> certain range (0 to 100). Then they are to be placed in an array
> without any duplications. However, the integers will have
> duplicates. So, these cannot be put into the array if the value is
> already in the array. Also, the order cannot be changed from the
> initial placement in the array.

Well, that's an unfortunate combination of requirements, since it
means you cannot use the many nice containers and algorithms designed
to make this problem easier, for example std::set, std::sort,
std::set_inters ection.

Yes you can. Here is the solution using "many nice containers and
algorithms"

Right. I should have been more specific, i.e., you can't use them
directly as your data structure.

I quite liked your sample code, except for building up a new set
containing the same old elements every time you add a new element. But
perhaps that's the safest way to go, and at only 100 elements it's
probably fast enough.

I'm not sure what you mean by that. Only one set is ever built...

Another idea would be to use a vector<bool>( 100 )... :-)

--
Magic depends on tradition and belief. It does not welcome observation,
nor does it profit by experiment. On the other hand, science is based
on experience; it is open to correction by observation and experiment.

[Neil Cerutti]

On 2006-02-27, Daniel T. <po********@ear thlink.net> wrote:

In article <sl************ **********@FIAD 06.norwich.edu> ,
Neil Cerutti <le*******@emai l.com> wrote:

I quite liked your sample code, except for building up a new set
containing the same old elements every time you add a new
element. But perhaps that's the safest way to go, and at only 100
elements it's probably fast enough.

I'm not sure what you mean by that. Only one set is ever built...

Er..., well, the pink elephants must have told me it. Sorry for the
mistaken reading.

--
Neil Cerutti
For those of you who have children and don't know it, we have a
nursery downstairs. --Church Bulletin Blooper