By Jean J Labrosse; et al
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Knowing how a language stores information can also be proactive in the optimization process, in that it gives the designer insight into the consequences of using a particular data type as it applies to storage and access overhead. This information may allow the designer to choose wisely enough to avoid the need for custom routines altogether. The following sections covering data structures will try to answer all five of these questions as they apply to each of the different data structures. 1 Simple Data Types The term “simple data type” refers to variables that store only one instance of data and store only one format of data at a time.
As an example, performing a multiply on a 24-bit LONG variable can use up 12 bytes of data storage just for the temporary storage of the upgraded term variables. So, it is generally advisable to resort to either an array of CHARs or, in extreme cases, an array of INTs to store large data values. This allows the designer to more tightly regulate the amount of data storage required. It also limits the amount of temporary data storage required for math, even though it will require a custom, and somewhat complicated, math routine.
2. Is there data present to be retrieved? 3. Is the buffer is full? 35 So, as the transmitter places data in the buffer, the storage pointer moves forward through the block of data memory. And as the receiver retrieves data from the buffer, the retrieval pointer chases the storage pointer. To prevent the system from running out of storage, both pointers are designed to “wraparound” to the start of the data block when they pass the end. When the protocol is operating normally, the storage pointer will jump ahead of the retrieval pointer, and then the retrieval pointer will chase after it.