Today we continue  discussing the best practice from storage engineering

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.

151) Querying: Storage products are not required to form a query layer. Still they may participate in the query layer on enumeration basis. Since the query layer can come separately over the storage entities, it is generally written in the form of a web service. However SQL like queries may also be supported on the enumerations as shown here: https://1drv.ms/w/s!Ashlm-Nw-wnWt1QqRW-GkZlfg0yz

152) Storage product code is usually a large code base. Component interaction diagrams can be generated and reorganizations may be done to improve the code.  The practice associated with large code bases applies to storage server code as well.

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