Today we continue  discussing the best practice from storage engineering:

146) Footprint: The code for a storage server can run on any device. Java for example runs on billions of devices and a storage server written in Java can run even on pocket devices. If the storage is flash and the entire storage server runs only on flash, it makes a great candidate for usability.

147) Editions: Just like the code for storage server can be made suitable for different devices, it can ship as different editions. One way to determine the different kinds of editions is to base it on where the customers demand it. Although there are many perspectives in these decisions, the ultimate service of the product is for the customer.

148) Standalone mode: Most storage products offer capabilities in a standalone mode. This makes it easy to try out the product without dependencies. One Box deployments also work in this manner. When we remove the dependency on the hardware, we enable the product to be easier to study and try out the features.

149) End User License Agreement:  Products have EULAs which the user is required to accept. If the EULA conditions are not read or accepted, certain functionalities of the server are turned off. Typically, this check happens at the initialization time of the product or the restart of the service. However, the initialization code is global and has very sensitive sequence of steps. Therefore, placing the EULA criteria without leaving the product in a bad state is necessary.

150) Provisioning: Storage artifacts are not just created by user. Some may be offered as built-ins so that the user can start working with their data rather than handling the overhead of setting up containers and describing them. These ready to use built-in storage artifacts lets the user focus on their tasks rather than the system tasks. A judicious choice of all such artifacts and routines are therefore useful to the user.

#codingexercise
int getCountOdd(node* root)
{
if (root == null) return 0;
int count;
while (root)  {
if (root->value %2 == 1) {
    count++;
}
root = root->next;
}
return count;
}