PDA

View Full Version : c++ concept i cant grasp



phinfanfrommd
10-19-2006, 01:21 AM
how would I find the smallest and largest numbers of an array of 10 integers? take it easy on me im no expert.

King Felix
10-19-2006, 01:24 AM
i'm lost :confused:

phinfanfrommd
10-19-2006, 01:27 AM
do you know any c++?

Nappy Roots
10-19-2006, 01:28 AM
its ****in simple

phinfanfrommd
10-19-2006, 01:29 AM
thanks i know it is
im just starting out give me a break

phinfanfrommd
10-19-2006, 01:29 AM
i dont need teh actualy code just the concept

Nappy Roots
10-19-2006, 01:32 AM
thanks i know it is
im just starting out give me a break



i was kiddin, i have noooooooo clue what your talkin about.

Dol-Fan Dupree
10-19-2006, 02:44 AM
off the top of my head a really simple solution would be have an integer value high and an integer value low. Set the low to the highest integer value and the high to the lowest. Then travse the array with a for loop. Like from i = 10 (or MAX_SIZE); i = 0; i++

Then check if a[i] > high then high equals i and also check if a[i] < low then low equals i.

There is probably a better way to do it, but I am too tired to think about it.

ChrisKo
10-19-2006, 12:22 PM
off the top of my head a really simple solution would be have an integer value high and an integer value low. Set the low to the highest integer value and the high to the lowest. Then travse the array with a for loop. Like from i = 10 (or MAX_SIZE); i = 0; i++

Then check if a[i] > high then high equals i and also check if a[i] < low then low equals i.

There is probably a better way to do it, but I am too tired to think about it.

That would work.

You could also use any of the sorting routines (bubble, quick, btree, etc) and once the array has been sorted, the first and last elements are your needed values.

phinfanfrommd
10-20-2006, 12:19 AM
That would work.

You could also use any of the sorting routines (bubble, quick, btree, etc) and once the array has been sorted, the first and last elements are your needed values.

wow that is much more efficient, im looking at it in my book now, sounds a little to complex for my simple solution. thanks though!