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# LAB 3 Arrays and Strings

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LAB 3 Arrays and Strings, Relational and Logic operators,
Type conversion, Bitwise operations

Problem A Arrays and Strings (cont.)
This and the next question walk you through more exercises on manipulating arrays and strings.
Arrays are so important in C that we will deal with them throughout the C part of the course.
Download lab3.c. This short program uses an array of size 12 to store input strings read in
using scanf, and simply outputs the array elements (char and its index) after each read
(similar to the last part of lab2D.c of lab2).
First observe the initial values of the array. Arrays without explicit initializer get random values.
Run it again and you might see the change of the strange values. This implies that when you
manually store characters into a char array to create a string, don’t assume that the uninitialized
array was filled with \0s. You need to manually add a \0 after the last stored character.
Now enter helloworld, observe that the array is now stored as
where \0 is added to the end of the input
texts, and ? is \0 or a random value. printf(“%s”) prints helloworld, with size 12
and length 10.
Next, enter a shorter word such as good, then observe that the array is stored as
and thus printf(“%s”) prints good
with size 12, length 4.
Next, enter hi, then observe that the array is store as
and printf(“%s”) prints hi with
size 12, and length 2. Now enter a word that is longer than hi (such as 01234567), see
what happens. Also as in lab2, you can manually add \0 into the array. If your \0 is before the
first existing \0, you essentially create a shorter string.
The key point here is that when an array is used to store a string, not all array elements got
reset. Thus, when you enter hello, don’t assume that the array contains character h e l o
and \0 only – there may exist random values, there may also exist characters from previous
storage. So it is always critical to identify the first \0 encountered when scanning from left to
right, ignoring characters thereafter. As observed, string manipulation library functions, such as
printf(“%s”), strlen, strcpy, strcmp follow this rule: scan from left to right,
terminate after encountering the first \0 character. Your string related functions should follow
the same rule.
Now enter quit , and observe that the program does not terminate. As discussed in class,
comparing arrays/string using == will not compare the content of the arrays/strings. Press Ctrl+C
to terminate the program “brutally”. You will explore different ways to compare string content
soon.
Finally, uncomment the statement str=’a’; and compile again. Observe that no
compiling error or warning is given. When running the program, it may or may not crash. So
unlike in Java, C compilers will not do array boundary checking.
No submission for problem 0.
h e l l o w o r l d \0 ?
g o o d \0 w o r l d \0 ?
h i \0 d \0 w o r l d \0 ?
2
Problem B0 Character arrays and strings (cont.)
Specification
Standard library defines a library function atoi. This function converts an array of digit
characters, which represents a decimal integer literal, into the corresponding decimal integer.
E.g., given a char array (string) s of “134”, which is internally stored as ‘1’ ‘3’ ‘4’ ‘\0’ ….. ,
function atoi(s) returns an integer 134.
Implement your version of atoi, call it my_atoi, which does exactly the same conversion.
Implementation
assumed to be a valid integer literal, the program first prints it as a string, and then call both
atoi and myatoi to convert it, and output its numerical value in decimal, hex and oct,
followed by double the value and square of the value. The program keeps on reading from the
user until quit is entered.
Complete the while loop in main(), and implement function my_atoi.
• Page 43 of the recommended K&R book “The C programing language” describes an
approach to convert a character array into decimal value, this approach traverses the array
from left to right. (You probably used a similar approach in lab2.)
A more intuitive approach, which you should implement here, is to calculate by traversing
the array from right to left, following the traditional concept ‘2’ ‘1’ ‘3’ ‘4’ ‘\0’ …..
Hint: the loop body you are going to write is different
from, and slightly more complicated than that in the
recommended textbook, but the logic is clearer (IMHO).
• For finding the right end of string, you can use your length() function you implemented
earlier. You can also explore the string library function strlen().
If you need, you can implement a helper function power(int base, int n) to
calculate the power. In next class we will learn to use math library functions. Don’t use Math
library function here.
• For detecting quit, since strings content cannot be compared directed. You can use the
isQuit() function you seen earlier, but you are also encouraged to explore the string
library function strcmp(). You can issue man strcmp to view the manual. Note that
strcmp()returns 0 (false) if the two argument strings are equal.
Sample Inputs/Outputs:
red 127 % a.out
Enter a word of positive number or ‘quit’: 2
2
atoi: 2 (02, 0X2) 4 4
my_atoi: 2 (02, 0X2) 4 4
Enter a word of positive number or ‘quit’: 4
4
atoi: 4 (04, 0X4) 8 16
my_atoi: 4 (04, 0X4) 8 16
Enter a word of positive number or ‘quit’: 9
9
atoi: 9 (011, 0X9) 18 81
my_atoi: 9 (011, 0X9) 18 81
103
102
101
100
3
Enter a word of positive number or ‘quit’: 12
12
atoi: 12 (014, 0XC) 24 144
my_atoi: 12 (014, 0XC) 24 144
Enter a word of positive number or ‘quit’: 75
75
atoi: 75 (0113, 0X4B) 150 5625
my_atoi: 75 (0113, 0X4B) 150 5625
Enter a word of positive number or ‘quit’: 100
100
atoi: 100 (0144, 0X64) 200 10000
my_atoi: 100 (0144, 0X64) 200 10000
Enter a word of positive number or quit: quit
red 128 %
No submission for this exercise.
Problem B Character arrays and strings (cont.) (20 pts)
Extend the program you developed above, so that myatio() function can convert any base of
2~10. (The library function atoi can only handle decimal literals correctly.) This version of
my_atoi takes two arguments: an integer literal and a base. Assume the input string is always a
valid integer literal of the specified base. For example, if the base is 2, the literal only contains 0
and 1. If the base is 5, the literals contains digit 0~4.
Note:
• Use scanf(“%s %d”,…) to read in a string followed by an integer of 2~10.
• The program terminates when use enter quit followed by any integer.
• Apparently, you should not call atoi() in my_atoi().
• In my_atoi(), you also should not call other functions declared in <stdlib.h>, such as
atol(), atof(), strtol(), strtoul().
• In my_atoi(), you also should not call library functions declared in <stdio.h>, such as
sscanf(), fscanf().
Sample Inputs/Outputs:
red 127 % a.out
Enter a word of positive number and base, or ‘quit’: 37 10
37
my_atoi: 37 (045, 0X25) 74 1369
Enter a word of positive number and base, or ‘quit’: 37 8
37
my_atoi: 31 (037, 0X1F) 62 961
Enter a word of positive number and base, or ‘quit’: 122 4
122
my_atoi: 26 (032, 0X1A) 52 676
Enter a word of positive number and base, or ‘quit’: 122 5
122
my_atoi: 37 (045, 0X25) 74 1369
4
Enter a word of positive number and base, or ‘quit’: 122 7
122
my_atoi: 65 (0101, 0X41) 130 4225
Enter a word of positive number and base, or ‘quit’: 1101 2
1101
my_atoi: 13 (015, 0XD) 26 169
Enter a word of positive number and base, or ‘quit’: 1001100 2
1001100
my_atoi: 76 (0114, 0X4C) 152 5776
Enter a word of positive number and base, or ‘quit’: 345 6
345
my_atoi: 137 (0211, 0X89) 274 18769
Enter a word of positive number and base, or ‘quit’: 345 8
345
my_atoi: 229 (0345, 0XE5) 458 52441
Enter a word of positive number and base, or ‘quit’: 3214 5
3214
my_atoi: 434 (0662, 0X1B2) 868 188356
Enter a word of positive number and base, or ‘quit’: 11111111 2
11111111
my_atoi: 255 (0377, 0XFF) 510 65025
Enter a word of positive number and base, or ‘quit’: quit 4
red 128 %
Submit your program using submit 2031A lab3 lab3myatoi.c
Problem C0 Increment and Decrement Operators
As discussed in class, C and other modern languages such as Java, C++ all support increment and
decrement operators ++ and –. These operators can be used as prefix or postfix operators,
appearing before or after a variable.
• the first two printf statements, one after x++ and one after ++x, both output 2. Do you
understand why? ++x does pre-increment, incrementing x ‘immediately’ , and x++ does
post-increment, incrementing x ‘later’ – at some point after the current statement but
before the next statement. Thus in both case when the printf statement is executed, x has
• Since post-increment x++ increments x ‘later’ – some point after the current statement
(and before the next statement), the two printf statement printf(“%d”, x++) and
printf(“%d”,++x) produce different results. In particular, since x++ increments x after
the current function call, printf(“%d”, x++) outputs the value before the increment
happens, whereas printf(“%d”, ++x) output the value after the increment happens.
In both cases, the printf statements after these two printf statements both output 2, as x
has been incremented at that point.
5
• When other operators such as assignment operators are involved, the result are also
different.
o y = x++ will assign y the un-incremented value of x, as x will be incremented after
the assignment statement.
o On the other hand, y = ++x will assign y the incremented value of x, as x is
incremented immediately, before the assignment is executed.
• By using ++ and — operators judiciously, code for traversing arrays can become succulent,
as shown in the last block of the code.
Download the Java version of program IncreDecre.java, compile and run it in
commandline, and observe the same result as in C (so in this course you will also improve your
Java skills too, as I have promised )
No submission for this question.
Problem C0 ‘Boolean’ in ANSI-C. Relational and logical operators
As discussed in class, ANSI-C has no type ‘Boolean’. It uses integers instead. It treats non-zero
value as true, and returns 1 for ture result. It treats 0 as false, and return 0 for false result.
• Observe that
o relational expression 3>2 has value 1, and 3<2 has value 0
o ! non-zero has value 0, !0 has value 1.
▪ Note that in Java, these are invalid expressions.
o && return 1 if both operands are non-zero, return 0 otherwise. || return 1 if either
operand is non-zero, and return 0 otherwise.
▪ Note that in Java, these are invalid expressions.
• Assume the author mistakenly used = , rather than ==, in three of the five if conditions.
Observe that although x has initial value 100, both if(x=4) and if(x=-2) clauses were
executed. This illustrates a few interesting things in ANSI-C:
o Unlike a Java complier, gcc does not treat this as a syntax error.
o Assignment expression such as x=4 has a return value, which is the value being
assigned to. So if(x=4) becomes if(4), and if(x=-2) becomes if(-2), and
if(x=0) becomes if(0)
o Any non-zero number is treated as ‘true’ in selection statement. Thus if(x=4) and
if(x=-2) are both evaluated to be true and their corresponding statements were
executed. On the other hand, 0 is treated as ‘false’, so if(x=0) was evaluated to be
false and its statement was not executed.
o Also observe that although if(x=0) condition was evaluated to be false, the
assignment x=0 was executed (before the evaluation) and thus x has value 0 after the
three if clauses.
• Observe that although the loop in the program intends to break when i becomes 8 and
thus should execute and prints 8 times, only hello 0 is printed. Look at the code for the
loop, do you see why? Fix the loop so that the loop prints 9 times, as shown below.
hello 0
hello 1
hello 2
6
hello 3
hello 4
hello 5
hello 6
hello 7
hello 8
No submissions for this exercise.
Problem C scanf, arithmetic and logic operators (20 pts)
Specification
Write an ANSI-C program that reads an integer from standard input, which represents a year,
month and day, and then determines how many days has elapsed in the year.
Implementation
• keep on reading a (4 digit) integer of year, followed by month and day, until a negative year
is entered (followed by any month and day).
• define a ‘Boolean’ function int isLeap(int year) which determines if year
represents a leap year. A year is a leap year if the year is divisible by 4 but not by 100, or
otherwise, is divisible by 400.
• implement function int countDays(int month, int day, int isLeap)
where month and day represent the current month and day of a year, and isLeap
indicates whether the year is a leap year. The function calculates how many days have
elapsed since the start of the year, including current day. There are 31 days in Jan, Mar,
May, July, Aug, Oct and Dec, and there are 30 days in April, June, Sep, and Nov. There are 29
days in Feb if the year is a leap year, and 28 days in Feb if the year is not a leap year.
• call the functions in main, and produce output as shown below. (the two functions do not
produce output). Note that if a year is leap year then the output ends with [leap year].
• put the definition (implementation) of your functions after your main function.
4.3 Sample Inputs/Outputs:
red 364 % gcc -Wall lab3Leap.c -o leap
red 365 % leap
Enter date (‘YYYY MM DD’): 2010 1 1
1 days of year 2010 have elapsed
Enter date (‘YYYY MM DD’): 2011 8 8
220 days of year 2011 have elapsed
Enter date (‘YYYY MM DD’): 2012 8 8
221 days of year 2012 have elapsed [leap year]
Enter date (‘YYYY MM DD’): 2400 8 8
221 days of year 2400 have elapsed [leap year]
Enter date (‘YYYY MM DD’): 2010 10 1
274 days of year 2010 have elapsed
Enter date (‘YYYY MM DD’): 2012 10 1
275 days of year 2012 have elapsed [leap year]
7
Enter date (‘YYYY MM DD’): 2100 11 4
308 days of year 2100 have elapsed
Enter date (‘YYYY MM DD’): 2032 11 4
309 days of year 2032 have elapsed [leap year]
Enter date (‘YYYY MM DD’): 2031 2 12
43 days of year 2031 have elapsed
Enter date (‘YYYY MM DD’): 2032 2 12
43 days of year 2032 have elapsed [leap year]
Enter date (‘YYYY MM DD’): -1 5 4
red 366 %
Submit your program by issuing submit 2031A lab3 lab3Leap.c
Problem D Type conversions in arithmetic, assignment, and
function calls (10 pts)
Specification
Write an ANSI-C program that reads inputs from the user one integer, one floating point
number, and a character operator. The program does a simple calculation based on the two
input numbers and the operator. The program continues until both input integer and floating
point number are -1.
Implementation
• download partially implemented program lab3conv.c, compile and run it. Observe that
o 9/2 gives 4, not 4.5. (When the result is converted to float, get 4.0).
o In order to get 4.5, we need to convert 9 or 2 (or both) to float, before division.
▪ One trick is to multiply 9 or 2 by 1.0. This forces the conversion from int to
float. Note that this must be done before the division.
▪ The “official” approach, is to explicitly cast 9 or 2 to float, using the cast
operator (float). Note that this must be done before the division. Cast after
the division does not work correctly.
o When assigning a float value to an int variable, the int variable gets the
integral part value. The floating point part is truncated (without any warning.) Also
no rounding occur here.
Note, the first two observations are the same in Java. For the last observation, when
assigning a float value to an int variable, Java will give compilation error because
“possible lossy conversion from float to int”. In this case, an explicit cast is required. (If you
try Java, even float f2 = 3.963; generates an complier error, why?)
• use scanf to read inputs (from Standard input), each of which contains an integer, a
character (‘+’, ‘-‘ ‘*’ or ‘/’) and a floating point number (defined as float)
separated by blanks. Assume all the inputs are valid.
• define a function float fun_IF (int, char, float) which conducts arithmetic
calculation based on the inputs
8
• define another function float fun_II (int, char, int) which conducts
arithmetic calculation based on the inputs
• define another function float fun_FF (float, char, float) which conducts
arithmetic calculation based on the inputs
• note that these three functions should have the same code in the body. They only differ in
the parameter type and return type.
• pass the integer and the float number to both the three functions directly, without explicit
type conversion (casting).
• display prompts and outputs as shown below.
• Once the program is running, observe the output of the first 2 lines, where conversions
happen in arithmetic and assignment operations. Convince yourself of the outputs (why
three arithmetic operations have different results, why i and j both get 3?)
Sample Inputs/Outputs:
red 330 % a.out
9/2=4.000000 9*1.0/2=4.500000 9/2*1.0=4.000000 9/(2*1.0)=4.500000
(float)9/2=4.500000 9/(float)2=4.500000 (float)(9/2)=4.000000
3.0*9/2/4=3.375000 9/2*3.0/4=3.000000 9*3/2*3.0/4=3.000000
i: 3 j: 3
Enter operand_1 operator operand_2 separated by blanks> 12 + 22.3024
Your input ’12 + 22.302401′ result in
34.302399 (fun_IF)
34.000000 (fun_II)
34.302399 (fun_FF)
Enter operand_1 operator operand_2 separated by blanks> 12 * 2.331
Your input ’12 * 2.331000′ result in
27.972000 (fun_IF)
24.000000 (fun_II)
27.972000 (fun_FF)
Enter operand_1 operator operand_2 separated by blanks> 2 / 9.18
Your input ‘2 / 9.180000’ result in
0.217865 (fun_IF)
0.000000 (fun_II)
0.217865 (fun_FF)
Enter operand_1 operator operand_2 separated by blanks> -1 + -1
red 331 %
Do you understand why the results of the fun-IF and fun-FF are same but both are different
from fun-II? Write a brief justification on the program file (as comments).
Submit your program using submit 2031A lab3 lab3conv.c
Problem E0 Bitwise operations
In class we covered bitwise operators & | ~ and << >>. It is important to understand that,
• A bit has value either 0 or 1. When using bitwise operator & |, value 0 is treated as False
and 1 is treated as True. Following the truth table of Boolean Algebra (True AND True is
9
True, False AND True is False etc.), for a bit b (which is either 0 or 1), there are 4
combinations.
▪ b & bit 0 generates a bit that is 0 (anything AND with False is False)
▪ b | bit 1 generates a bit that is 1 (anything OR with True is True)
▪ b & bit 1 generates a bit that is the same as b. (AND with True, no change)
▪ b | bit 0 generates a bit that is the same as b. (OR with False, no change)
• each bitwise operation generates a new value but does not change the operand itself. For
example, for an int variable abc, expression abc <<4, abc & 3, abc | 5 does not
change abc. In order to change abc, you have to use abc = abc <<4, abc = abc & 3,
abc = abc | 5, or use their compound assignment versions abc <<= 4, abc &=3,
abc |= 5. When these expressions are executed, based on the above observations, we got
the following idioms:
▪ b = b & bit 0 sets b to 0 (“turns bit b off”),
▪ b = b | bit 1 sets b to 1 (“turns bit b on”),
▪ b = b & bit 1 sets b to its original value (“keep the value of b”).
▪ b = b | bit 0 sets b to its original value (“keep the value of b”).
several bitwise operations. It terminates when -1000 is entered.
Compile and run the program with several inputs, and observe
• what the resulting binary representations look like when the input abc is left bit shifted, and
is bit flipped. Note that expression abc << 3 or ~abc does not modify abc itself, so the
program uses the original value in other operations.
• how 1 << 4 is used with | to turn on bit-4 (denote the right-most bit as bit-0). Again,
expression abc | 1<<4 does not change abc itself.
As a C programming idiom (code pattern), for an integer abc, abc = abc |(1<<j) turns
on bit-j of abc, i.e., bit-j becomes 1 regardless of its original value (other bits remain
unchanged).
• what the bit representation of ~(1<<4) looks like, and how it is used with bitwise operator
& to turn off bit 4. As a C programming idiom here, abc = abc & ~(1 << j) turns off
bit-j of abc, i.e., bit-j becomes 0 regardless of its original value (other bits remain
unchanged).
Also observe here that parenthesis is needed around 1<<4 because operator << has lower
precedence than operator ~. (What is the result of ~1<<4 ?)
• how 1 << 4 is used with & to keep bit 4 and turn off all other bits. As a programming
idiom, if (abc & (1<<j)) is used to test whether bit-j of abc is on (why?).
• what the bit representation of 077 looks like, and how it is used with & to keep the lower 6
bits and turn off all other bits. Change 077 to 0177 and then 0377 to see what happens.
• what the bit representation of ~077 looks like, and how it is used with & to turn off lower 6
bits and keep all other bits.
Enter different numbers, trying to understand these bitwise idioms.
When you terminate the loop, observe the for unsigned int 0xFFFFFFFF, whose binary
representation is 11…111, left shift >> 3 add 000 on the left end, as specified in C. For signed
int 0xFFFFFFF, C did not specify if 000 or 111 will be added in and it is implementationdependent. If you run the program in the lab, it shows that 111 is added, but this maybe
different in other platforms. So the rule of thumb is: In C, don’t do right shift on signed integers!
As mentioned in class, Java introduced >>> to resolve the ambiguity.
10
Finally, observe that bitwise operation is used in function printBinary() to generate artificial
‘0’ or ‘1’, printing the binary representation of an int. Try to understand the code.
No submission for this question, but doing the exercise gets you prepared for problems E1, E2
below.
What are Bitwise operators used for? The following two questions walk you through two
applications of bitwise operations.
Problem E1 bits as Boolean flags (20 pts)
Specification
In class we mentioned that one usefulness of bitwise operator is to use bits as Boolean flags.
Here is an example. Recall that in lab 2 we have the problem of counting the occurrence of digits
in user inputs. We used an array of 10 integers where each integer element is a counter. Now
consider a simplified version of the problem: there is no need to count the number of
occurrences of each digit, instead we just need to record whether each digit has appeared in the
input or not (no matter how many times they appear). For example, for input EECS2031A,
2019-21FW, LAS0006, we need to record that 0,1,2,3, 6 and 9 appeared in the inputs, but
4,5,7 and 8 did not. One way to do this is to maintain an array of 10 (short) integers, where
each integer element is used as a Boolean flag: 0 for False (absent) and 1 for True (present).
Now imagine in old days when memory is very limited, and thus instead of 10 integers, which
takes 160~320 bits, you can only afford to use one integer (16~32 bits) to do the job. Is it
possible?
Here the bitwise operations come to the rescue. The idea is that since we only need a True/False
info for each digit, 1 bit is enough for each digit, so we need only a total of 10 bits to record.
Thus an integer or even a short integer is enough. Specifically, we declare a short int
variable named flags, which usually has 16 bits. Then we designate 10 bits in flags as
Boolean flags digits 0~9. For example, we designate the right most bit (denoted bit-0) as the
Boolean flag for digit 0, designate the next bit (denoted bit-1) as the Boolean flag for digits 1,
and so on. flags is initially set to 0. Then after reading the first digit, say, 2, we use bitwise
operation to “turn on” (set to 1) bit-2 of flags. So flags’ bit representation becomes
00000000 00000100. Later when reading another 2, you can somehow check if bit-2 is on and
turns it on if not, or alternatively, simply use the same operation to turn on bit-2 of flags,
although bit-2 is already on. After reading all inputs EECS2031A, FW2019-21, LAS1006A,
which contains digit 0,1,2,3,6 and 9, the internal representation of flags becomes
0000010 01001111. That is, bit 0,1,2,3,6 and 9 are on. Finally, we can use bitwise operations
to examine the lower 10 bits of flags, determining which are 1 and which are 0.
Implementation
on reading inputs using getchar until end of file is entered. It then outputs if each digits is
present in the inputs or not.
• Observe that by putting getchar in the loop header, we just need to call getchar once.
(But a parenthesis is needed due to operator precedence).
• Complete the loop body, using one of the idioms mentioned on previous page, so that
flags is updated properly after reading a digit char.
• Complete the output part, by checking the right most 9 bits one by one. Hint: there are at
least two approaches to check whether a particular bit is 1 or 0. One of the idiom mention
earlier can do this, another approach is hinted in the printBinary function provided.
11
• For your convenience, a function printBinary() is defined and used to output the binary
representation of flags, both before and after user inputs. (This is the same function given
in lab3bit.c)
It is interesting to observe that function printBinary() itself uses bitwise operations to
generate artificial ‘0’ or ‘1’ , printing the binary representation of an int. It would be
interesting to trace the code. This may help you complete the output part.
red 369 % a.out
flags: 00000000 00000000
YorkU LAS C
^D (press Ctrl and D)
flags: 00000000 00000000
0: No
1: No
2: No
3: No
4: No
5: No
6: No
7: No
8: No
9: No
red 370 % a.out
flags: 00000000 00000000
EECS2031A FW2019-21
LAS1006A
^D (press Ctrl and D)
flags: 00000010 01001111
0: Yes
1: Yes
2: Yes
3: Yes
4: No
5: No
6: Yes
7: No
8: No
9: Yes
red 371 % a.out
flags: 00000000 00000000
EECS3421 this is good 3
429Dk
^D (press Ctrl and D)
flags: 00000010 11111111
0: Yes
1: Yes
2: Yes
3: Yes
4: Yes
12
5: Yes
6: Yes
7: Yes
8: No
9: Yes
red 372 % a.out < input2C.txt
flags: 00000000 00000000
flags: 00000000 11111111
0: Yes
1: Yes
2: Yes
3: Yes
4: Yes
5: Yes
6: Yes
7: Yes
8: No
9: No
red 373 %
Submit your program by issuing submit 2031A lab3 lab3flags.c
Problem E2 Bitwise operation (20 pt)
Specification
A digital image is typically stored in computer by means of its pixel color values, as well as some
formatting information. Each pixel color value consists of 3 values of 0 ~ 255, representing red
(R), green (G) and blue (B).
As mentioned in class, Java’s BufferedImage class has a method int getRGB(int x,int y),
which allows you to retrieve the RGB color value of an image at pixel position (x,y). How could
the method return 3 values at a time? Instead of returning an int array or an object, the ‘trick’ is
to return an integer (32 bits) that packs the 3 values into it. Since each value is 0~255 thus 8 bits
are enough to represent it, a 32 bits integer has sufficient bits. They are packed in such a way
that, counting from the right most bit, B values occupies the first 8 bits (bit 0~7), G occupies the
next 8 bits, and R occupies the next 8 bits. This is shown below. (The left-most 8 bits is packed
with some other information about the image, called Alpha, which we are not interested here.)
7 6 5 4 3 2 1 0 7 6 5 4 3 2 1 0 7 6 5 4 3 2 1 0 7 6 5 4 3 2 1 0
Suppose a pixel has R value 18 (which is binary 00010010), G value 7 (which is binary 00000111)
and B value 8 (which is binary 00001000), and Alpha has value 100, then the integer is packed as
0 0 1 0 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0
Implementation
In this exercise, you are going to use bitwise operations to pack input R,G and B values into an
integer, and then use bitwise operations again to unpack the packed integer to retrieve the R, G
and B values.
100 for Alpha 18 7 8
31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
0
13
Each input contains 3 integers representing the R, G and B value respectively, and then outputs
the 3 values with their binary representations (implemented for you). The binary
representations are generated by calling function void printBinary(int val), which is
defined for you in another program binaryFunction.c.
Next is the packing part that you should implement. This packs the 3 input values, as well as
Alpha value which is assumed to be 100, into integer variable rgb_pack.
Then the value of rgb_pack and its binary representation is displayed (implemented for you).
Next you should unpack the R, G and B value from the packed integer rgb_pack. Note that
you should not just use the original R,G and B value, as this is considered a “hoax”.
After that, the unpacked R,G and B value and their Binary, Octal and Hex representations are
displayed (implemented for you).
The program terminates when you enter a negative number for either R, G or B value.
Hint: Packing might be a little easier than unpacking. Considering shifting R,G,B values to the
proper positions and then somehow merge them into one integer (using bitwise operators). For
unpacking, you can either do shifting + masking, or, masking + shifting, or, shifting only. Shifting
+ masking means you first shift the useful bits to the proper positions, and then turn off (set to
0) the unwanted bits while keeping the values of the useful bits. What you want to end up with,
for example for R value, is a binary representation of the following, which has decimal value 18.
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0
Masking + shifting means you first use & to turn off some unrelated bits and keep the values of
the good bits, and then do a shifting to move the useful bits to the proper position. When doing
shifting, the rule of thumb is to avoid right shifting on signed integers. Explore different
approaches for unpacking.
Finally, it is interesting to observe that function printBinary() itself uses bitwise operations
to generate artificial ‘0’ or ‘1’, printing the binary representation of the int.
Since printBinary() is defined in another file, how to use a function that is defined in
another file? Don’t do include<binaryFunction.c>, instead, declare the function
printBinary() in the main program lab3RGB.c (how?), and then compile the files together,
as shown below. (We will talk more about multiple files next week).
Sample Inputs/Outputs:
red 338 % gcc lab3RGB.c binaryFunction.c
red 339 % a.out
enter R value (0~255): 1
enter G value (0~255): 3
enter B value (0~255): 5
A: 100 binary: 00000000 00000000 00000000 01100100
R: 1 binary: 00000000 00000000 00000000 00000001
G: 3 binary: 00000000 00000000 00000000 00000011
B: 5 binary: 00000000 00000000 00000000 00000101
Packed: binary: 01100100 00000001 00000011 00000101 (1677787909)
Unpacking ……
R: binary: 00000000 00000000 00000000 00000001 (1,01,0X1)
G: binary: 00000000 00000000 00000000 00000011 (3,03,0X3)
pay attention to how the
program is compiled
14
B: binary: 00000000 00000000 00000000 00000101 (5,05,0X5)
————————————
enter R value (0~255): 22
enter G value (0~255): 33
enter B value (0~255): 44
A: 100 binary: 00000000 00000000 00000000 01100100
R: 22 binary: 00000000 00000000 00000000 00010110
G: 33 binary: 00000000 00000000 00000000 00100001
B: 44 binary: 00000000 00000000 00000000 00101100
Packed: binary: 01100100 00010110 00100001 00101100 (1679171884)
Unpacking ……
R: binary: 00000000 00000000 00000000 00010110 (22, 026, 0X16)
G: binary: 00000000 00000000 00000000 00100001 (33, 041, 0X21)
B: binary: 00000000 00000000 00000000 00101100 (44, 054, 0X2C)
————————————
enter R value (0~255): 123
enter G value (0~255): 224
enter B value (0~255): 131
A: 100 binary: 00000000 00000000 00000000 01100100
R: 123 binary: 00000000 00000000 00000000 01111011
G: 224 binary: 00000000 00000000 00000000 11100000
B: 131 binary: 00000000 00000000 00000000 10000011
Packed: binary: 01100100 01111011 11100000 10000011 (1685840003)
Unpacking ……
R: binary: 00000000 00000000 00000000 01111011 (123, 0173, 0X7B)
G: binary: 00000000 00000000 00000000 11100000 (224, 0340, 0XE0)
B: binary: 00000000 00000000 00000000 10000011 (131, 0203, 0X83)
————————————
enter R value (0~255): 254
enter G value (0~255): 123
enter B value (0~255): 19
A: 100 binary: 00000000 00000000 00000000 01100100
R: 254 binary: 00000000 00000000 00000000 11111110
G: 123 binary: 00000000 00000000 00000000 01111011
B: 19 binary: 00000000 00000000 00000000 00010011
Packed: binary: 01100100 11111110 01111011 00010011 (1694399251)
Unpacking ……
R: binary: 00000000 00000000 00000000 11111110 (254, 0376, 0XFE)
G: binary: 00000000 00000000 00000000 01111011 (123, 0173, 0X7B)
B: binary: 00000000 00000000 00000000 00010011 (19, 023, 0X13)
————————————
enter R value (0~255): -3
enter G value (0~255): 3
enter B value (0~255): 56
red 340 %
15
Assume all the inputs are valid.
Submit your program by issuing submit 2031A lab3 lab3RGB.c
Make sure your program compiles in the lab environment. The program that does
not compile in the lab will get 0.
All submissions need to be done from the lab.
In summary, for this lab you should submit:
lab3myatoi.c, lab3Leap.c, lab3conv.c, lab3flags.c, lab3RGB.c
At any time and from any directory, you can issue submit -l 2031A lab3 to see the list of
files that you have submitted.
Common Notes
All submitted files should contain the following header:
/***************************************
* EECS2031A – Lab3 *
* Author: Last name, first name *