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README for Homework 6 - UM By Nick Flores & Daniel Rodrigues We have received help from the following: Chase Jacob Everything has been correctly implemented. Departures: In our design we initially planned to complete the assignment in 3 distinct modules, but found we really only needed 2. One for our UM main (um.c) and one for memory segmentation (segments.c). Our third was going to be a module for the UM's I/O but we realized that wasn't necessary after understanding the assignment layout a bit more. We also planned to use Hanson Sets but ended up only using Sequences. Architecure: [ UM ] | | [ Segments ] The architecture of our system (rougly shown in the digram above) consists of the UM (main) and the Segments (memory) modules. Segments is seen here as a child of the UM module. Modules: um.c: -This is our main module which houses most of the UM's heavy lifting. -It reads in the binary file of instructions, loads the program, and executes the instruction cycle. -It also contains the implementation for each instruction opcode as well as the array of registers that is used to execute the relative instructions. -This module also creates an instance of our memory struct that is used across all functions/modules. -um.c's main secret is its array of registers, which are only modified in this module but whose values are passed to segments. -um.c is segments.c's parent, so only it has knowledge of the latter. segments.c: -This is our main memory module, which houses functions to manipulate and touch the segments in memory. -This module is used in order to load, store, map, unmap, and replace segments of memory. -segments.c makes excellent use of Hanson's Sequences' to represent the segments in memory, which are sequences of sequences. -The secret of segments.c is hiding how our UM deals with the segmented memory. -This is the child module of um.c. Execution Time: We ran midmark.um in approximately 4 seconds, which executes 8.5 million instructions, so basic arithmatic tells us that a 50 million instruction program would take about 24 seconds. UM Unit Tests: print-H.um: This test takes in an input and outputs that value immediately. It then loads 2 values into 2 separate registers, calculates the sum, and outputs the result, which should be 72 (or the character 'H'). load_prgm: This test loads values into registers and then tests the load progam function. If load program is nonfunctional the test will halt without outputting anything, otherwise it will output the character 'H'. conditional.um: This test loads values into registers and then tests the conditional move function. If functional it will outpu 'H', otherwise it will output nothing (or the NULL char). load_seg.um: This tests store and load for functionality. It maps a new segment, stores a value in that segment, retrieves that valut and outputs it. A correct UM will output 'H' otherwise nothing. arith_test.um: This tests our arithmatic functions (add, mulitply, divide, and nand). map.um: This tests verbose mapping by loading values, maps 5 new segments, stores and loads from those segments, then unmaps them and repeats after unmapping. This is to make sure no new segments are created and our UM reuses umapped segments. A correct UM will produce 2 'H' characters for this test. Hours: Analyzing: 3 hours Designing: 5 hours Solving: 15+ hours
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