Corning Community College
ENGR1050 C for Engineers
To create a program that can perform the task of a binary counter, in incrementing or decrementing fashion.
In “The C Book”, please read through Chapter 5.
It is your task to write a program that, upon accepting various pieces of input from the user, acts as a binary counter, displaying the resultant values in binary as from a starting to an ending progression.
Your program should:
Some additional points of consideration:
lab46:~/src/c4eng/bcf0$ ./bcf0 Enter starting value (0-1023): 12 Enter ending value (0-1023): 28 0000001100 0000001101 0000001110 0000001111 0000010000 0000010001 0000010010 0000010011 0000010100 0000010101 0000010110 0000010111 0000011000 0000011001 0000011010 0000011011 0000011100 lab46:~/src/c4eng/bcf0$
The execution of the program is short and simple- obtain the input, do the processing, produce the output, and then terminate.
lab46:~/src/c4eng/bcf0$ ./bcf0 Enter starting value (0-1023): 17 Enter ending value (0-1023): 0 0000010001 0000010000 0000001111 0000001110 0000001101 0000001100 0000001011 0000001010 0000001001 0000001000 0000000111 0000000110 0000000101 0000000100 0000000011 0000000010 0000000001 0000000000 lab46:~/src/c4eng/bcf0$
lab46:~/src/c4eng/bcf0$ ./bcf0 Enter starting value (0-1023): 5543 ERROR: input value must be between 0-1023! lab46:~/src/c4eng/bcf0$
lab46:~/src/c4eng/bcf0$ ./bcf0 Enter starting value (0-1023): 255 Enter ending value (0-1023): 7168 ERROR: input value must be between 0-1023! lab46:~/src/c4eng/bcf0$
In the C4ENG public directory, inside the bcf0 subdirectory, will be a copy of my implementation (in executable form), which abides by the project specifications. Please compare its output against that of your implementation.
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:
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:
The binary for 15 (decimal) is 1111 (binary)
Or:
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.
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:
value = number & 128;
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.
Depending on the approach taken, you might have a need to store each of the bits for later display (for example, if you produce the bits in reverse order to the manner you wish to display them).
An array is like a regular variable, although instead of being able to store exactly one value, it acts as a container for MANY variables (all of the same type). We can then address each value through an offset (0 being the first slot, 1 being the second, etc.)
First, we need to identify a need; just as we needed to do with loops.
Let's say we had the following standalone variables:
int num1 = 13; int num2 = 73; int num3 = 26; int num4 = 57;
We likely understand how to work with each of the four independent variables, but we can't exactly automate accessing them, such as through a loop.
This is where an array can come in handy. Witness, the equivalent storage of numbers using an array:
int num[4]; // declare a 4 element integer array num[0] = 13; num[1] = 73; num[2] = 26; num[3] = 57;
What value does this offer us? Well, for one, we can automate the access of the array. Let's say we wanted to display the array contents (we have to do so one element at a time):
int index = 0; int max = 4; for (index = 0; index < max; index = index + 1) { fprintf (stdout, "%d\n", num[index]); }
Perhaps an array can be of some use in this project?
To successfully complete this project, the following criteria must be met:
To submit this program to me using the submit tool, run the following command at your lab46 prompt:
$ submit c4eng bcf0 bcf0.c Submitting c4eng project "bcf0": -> bcf0.c(OK) SUCCESSFULLY SUBMITTED
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:
78:bcf0:final tally of results (78/78) *:bcf0:proper error checking and status reporting performed [13/13] *:bcf0:correct variable types and name lengths used [13/13] *:bcf0:proper output formatting per specifications [13/13] *:bcf0:runtime tests of submitted program succeed [13/13] *:bcf0:no negative compiler messages for program [13/13] *:bcf0:code is pushed to lab46 repository [13/13]
Additionally: