Corning Community College
ENGR1050 C for Engineers
We continue our explorations of the electronics kit, by adding additional LEDs to control them in unison with a C program on our pi, having them count in binary.
Do note, the productive way to go about this project involves taking the following steps:
If you start too late, and do not ask questions, and do not have enough time and don't know what is going on, you are not doing the project correctly.
After exploring, assembling, and testing the intended circuit (4 LEDs), adapt the provided C code to use the bank of connected LEDs to count in binary from 0000 to 1111 (0 to 15).
Using the current value of count, your task is to make use of if statements and bitwise logic to determine from an ongoing count the state of the individual bits.
It is your task to write a C program that interfaces successfully with four independently connected LED circuits, arranged in some orientation to ascertain an order or positioning, where your program will (in endless fashion, or until being manually interrupted) display a count (in binary) of values from 0 to 15 (then rollover, or reset).
If “1” means the LED in that position is ON, and “0” means the LED in that position is OFF, then you want to write a program that performs the following progression (over and over again):
0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 <-- 15, the maximum value to display 0 0 0 0 <-- 0, we "roll over" and start again 0 0 0 1 0 0 1 0 ...
To assist with consistency across all implementations, data files for use with this project are available on lab46 via the grabit tool. Be sure to obtain it and ensure your implementation properly works with the provided data.
lab46:~/src/SEMESTER/DESIG$ grabit DESIG PROJECT
You will want to go here to edit and fill in the various sections of the document:
A logical operation that compares 2 operands and returns a 1 if both are the same or a 0 if they are not. In the case of binary numbers, the operations compares the 2 numbers digit by digit. For example, 0=0 would return 1, 1=1 would return 1, but 1=0 would return 0. In the case of 2 strings of numbers, like 111000 and 000000 the operator would return 000111 since the first through third digits match, but digits 4 through 6 are different.
Bitwise inclusive OR operator compared the bits of two operands and returns a value of “1” if the bit from either of the operands is a “1.” For example, comparing 1100110 and 0101010 would return 1101110. Looking at bother operands bit by bit and comparing them, any location that has at least one “1” will return a 1. This operator will also return a “1” if both operands contains a “1” at the same bit.
Similar to the bitwise inclusive OR operator, the bitwise exclusive OR operator will also compare the individual values of 2 operands bit-by-bit to compare the. However, unlike the inclusive variety, the exclusive OR will only return a 1 is ONE OF THE TWO BITS CONTAINS 1. For example, comparing 10101 and 11111 would return 01010.
If statements can be used to run a program by presenting the computer with a choice (or decision), as long as the “condition” the “if” is referring to is “true.” For example, if you used the statement: if (score⇐50), then the computer will run the code that follows it in the “box” if the data it is connected to does possess a “score” of below or equal to 50. If the “score” is higher than 50, the program will respond with an error. If this erroneous function is followed by another if statement, say “if (!score⇐50) the computer will present the data should the initial statement be “false.”
Another way to have the data presented, should there be an error, is to use “else.” If the statement doesn't match or agree with what the computer thinks of as “true,” the “else statement will allow the data to be presented as long as another statement is written. For example, if the statement of “if (condition) doesn't compute, adding a box after the first box starting with the command “else” followed by a different (condition), the computer will present the data referred to by the “else” command after skipping the “false” “if” statement.
Note: The more “if” statements you make in a program, the longer it will take for the computer to run the instructions, so it is best to have “else” statements ready, as the computer will merely skip any “false” “if” or “else” statements until it comes upon a “true” statement in the worst-case scenario, and ignore them should the first “if” statement be true.
The general flow of the process (one way of going about it, anyway) can be described as follows:
SET COUNTER TO ZERO
REPEAT:
SHOULD THE ONE'S POSITION HAVE A ONE:
ACTIVATE THE ONE'S PLACE
OTHERWISE:
DEACTIVATE THE ONE'S PLACE
SHOULD THE TWO'S POSITION HAVE A ONE:
ACTIVATE THE TWO'S PLACE
OTHERWISE:
DEACTIVATE THE TWO'S PLACE
SHOULD THE FOUR'S POSITION HAVE A ONE:
ACTIVATE THE FOUR'S PLACE
OTHERWISE:
DEACTIVATE THE FOUR'S PLACE
SHOULD THE EIGHT'S POSITION HAVE A ONE:
ACTIVATE THE EIGHT'S PLACE
OTHERWISE:
DEACTIVATE THE EIGHT'S PLACE
PAUSE FOR HUMAN PERCEPTION
LET THE COUNTER BE INCREMENTED BY ONE
To be successful in this project, the following criteria (or their equivalent) must be met:
Let's say you have completed work on the project, and are ready to submit, you would do the following (assuming you have a program called uom0.c):
lab46:~/src/SEMESTER/DESIG/PROJECT$ make submit
You should get some sort of confirmation indicating successful submission if all went according to plan. If not, check for typos and or locational mismatches.
I'll be evaluating the project based on the following criteria:
78:stl1:final tally of results (78/78) *:stl1:used grabit to obtain project by the Sunday prior to duedate [13/13] *:stl1:clean compile, no compiler messages [13/13] *:stl1:program conforms to project specifications [39/39] *:stl1:code tracked in lab46 semester repo [13/13]