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--- haas:fall2019:c4eng:projects:epf1 [2019/09/30 12:48] – created wedge
+++ haas:fall2019:c4eng:projects:epf1 [2019/09/30 19:08] (current) – wedge
@@ Line 11: / Line 11: @@
 =====Reading=====
 In "The C Book", please read through Chapter 5.
+=====Diagram=====
+To refresh your memory, here is a diagram of the circuit you can build to drive the LEDbar:
+{{:haas:fall2019:c4eng:projects:ledbar.png?400|}}
+Be sure to use 220 Ohm resistors for this.
 =====Program=====
@@ Line 73: / Line 81: @@
 lab46:~/src/c4eng/epf1$
 </cli>
+=====Obtaining binary values=====
+You might wonder how, when you are limited to non-binary input, how you can obtain the binary value so that you can work with it.
+There are a couple ways to go about this:
+====Method one: convert from decimal to binary====
+Using the division method, you can convert a decimal value to binary, continually dividing the value (or its quotient) by the base, until the quotient is 0, then we use the remainders to give us the binary value:
+  * value / base
+  * value = 15
+  * value / 2
+    * ie: 15 / 2
+      * quotient: 7
+      * remainder: 1
+  * value = quotient
+  * value / 2
+    * ie: 7 / 2
+      * quotient: 3
+      * remainder: 1
+  * value = quotient
+  * value / 2
+    * ie: 3 / 2
+      * quotient: 1
+      * remainder: 1
+  * value = quotient
+  * value / 2
+    * ie: 1 / 2
+      * quotient: 0
+      * remainder: 1
+The binary for 15 (decimal) is 1111 (binary)
+Or:
+  * value = 11
+  * value / 2
+    * quotient: 5
+    * remainder: 1
+  * value = 5
+  * value / 2
+    * quotient: 2
+    * remainder: 1
+  * value = 2
+  * value / 2
+    * quotient: 1
+    * remainder: 0
+  * value = 1
+  * value / 2
+    * quotient: 0
+    * remainder: 1
+The binary for 11 (decimal) is 1011 (binary).
+NOTE that the order in which we get the remainders produces the number from right to left.
+====Method 2: bitwise AND the place values====
+If one understands the weight values corresponding with the places of each bit in a binary number, we can simply do a bitwise AND and see if the result is greater than 0 or not:
+^  2 to the 7  ^  2 to the 6  ^  2 to the 5  ^  2 to the 4  ^  2 to the 3  ^  2 to the 2  ^  2 to the 1  ^  2 to the 0  ^
+|  128  |  64  |  32  |  16  |  8  |  4  |  2  |  1  |
+So, if we wanted to see if the 2 to the 7th bit is active, we can simply:
+<code c>
+    value  = number & 128;
+</code>
+We can then check of value is greater than 0; if it is, we've got a 1 in that position, if it isn't, we have a 0.
+We can then repeat the operation for 64, 32, 16, etc. down to 1, to get each bit of our number.
+NOTE that I have only taken us out to 8-bits. You may need to extend this to incorporate all the allowed values for this project.
 =====Submission=====
@@ Line 97: / Line 177: @@
 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.
-What I'll be looking for:
-<code>
-:epf1:final tally of results (78/78)
-*:epf1:proper error checking and status reporting performed [13/13]
-*:epf1:correct variable types and name lengths used [13/13]
-*:epf1:proper output formatting per specifications [13/13]
-*:epf1:runtime tests of submitted program succeed [13/13]
-*:epf1:no negative compiler messages for program [13/13]
-*:epf1:code is pushed to lab46 repository [13/13]
-</code>
-Additionally:
-  * Solutions not abiding by spirit of project will be subject to a 25% overall deduction
-  * Solutions not utilizing descriptive why and how comments will be subject to a 25% overall deduction
-  * Solutions not utilizing indentation to promote scope and clarity will be subject to a 25% overall deduction
-  * Solutions not organized and easy to read are subject to a 25% overall deduction

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