EECS2031 Lab 2 Solved

LAB 2: Character and integer literals, number systems, array and character arrays, operators

0 Problem 0 Number bases

• Visit the website www.cleavebooks.co.uk/scol/calnumba.htm to play with different bases of integer representations. Enter a valid number in any base and you will see the representations in all the other bases. For this course, we focus on binary, octal and hexadecimal representations. The [] beside each base indicates the valid symbols for the base. For example, valid symbols for base 8 are [0,1,2,3,4,5,6,7] whereas valid symbols for base 16 are [0…9,A,B,C,D,E,F]. Try entering 38 in base 8 and 4H in base 16 and observe how the calculator refuses to work for you. Enter 37 in base 8, or 31 in base 10 and calculate it. Convince yourself that the results in binary, octal and hexadecimal do have decimal value 31.
• To see how a 32 bits binary representation looks like, and also how negative numbers are represented in binary, please download, compile and run the Java program Binary.java. This program converts integer input into its 32 bits binary representation, which is the way integers are stored internally in memory. All inputs should be decimal integer literals. You can enter both positive and negative numbers (so you see how negative number is represented using 2’s complement). Stop the program by entering -10000.
• Download, compile and run the C program binaryFun0.c. This program reads in a decimal integer literal, and then outputs the integer in Decimal, Octal, and Hex. Look at the code and observe that Decimal, Octal and Hex can be displayed using printf with specific conversion specifications %d, %o and %X(or %x) respectively.
  Displaying binary representation, however, is not directly supported in printf, as C does not support integer literal in binary. In the next lab you will see an enhanced version of the program, which uses bitwise operations to display binary in C. Bitwise operations will be covered in lecture 3.
  All inputs should be decimal integer literals. Stop the program by entering -10000.

  No submissions for problem 0.

  1. Problem A Character literals

  1.1 Specification

  Write an ANSI-C program that reads input from the Standard input, and then outputs the processed information on the Standard output.
  Each user input contains an integer, followed by a blank and then a character. If the character does represent a digit, e.g., ‘3’, (how to check?), then the program outputs the sum of the integer and the numerical value that this character represents (how to get the numerical value of a character such as ‘3’?). If the character does not represent a digit, e.g., ‘A’, the program outputs that the character is not a digit.

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The program continues until the input integer is -10000 (and is followed by a blank and any one character).

1.2 Implementation

• name your program lab2A.c
• keep on reading and processing input, until an integer -10000 is read.
• use scanf (“%d %c”, ..) to read inputs.
• define a ‘Boolean’ function isDigit(char c) to determine if c represents a digit. We

  mentioned in class that ANSI-C does not have a type `boolean’, instead it uses 0 to represent false, and use non-zero integer to represent true. So, as a general practice in C, your function should return a non-zero number (usually 1) if c is a digit and returns 0 otherwise.
  Note that you should NOT use if c==’0′ c==’1′ c==’2′ c==’3′ … c==’9′. Instead, use the one-statement trick that we discussed in class.

• Note that in getting the numerical value of a digit character such as ‘3’, you should NOT use if c==’0′ c==’1′ c==’2′ c==’3′ … c==’9′. Instead, use the one- statement trick that we discussed in class.
• put the definition (implementation) of function isDigit after your main function.
• display the prompt and output as shown below.

  1.3 Sample Inputs/Outputs: (ONE blank line between each interaction/iteration): red 338 % gcc –Wall lab2A.c -o lab2a
  red 339 % lab2a
  Enter an integer and a character separated by blank>12 c Character ‘c’ does not represent a digit

  Enter an integer and a character separated by blank>12 9 Character ‘9’ represents a digit. Sum of 12 and 9 is 21

  Enter an integer and a character separated by blank>100 8 Character ‘8’ represents a digit. Sum of 100 and 8 is 108

  Enter an integer and a character separated by blank>120 ! Character ‘!’ does not represent a digit

  Enter an integer and a character separated by blank>-10000 a red 340 %

  Submit your program by issuing submit 2031Z lab2 lab2A.c 2. Problem B character literals

  2.1 Specification

  Write an ANSI-C program that uses getchar and putchar to read from the Standard input, and outputs (duplicates) the characters to the Standard output. For each input character that is a lower-case letter (how to check?), converts it into the corresponding upper case letter in the output (how to convert?). The program continues to read and output until EOF is entered.

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2.2 Implementation

You might want to start with the template file copy.c provided for lab1.

• name your program lab2B.c
• use getchar() and a loop to read characters.
• use putchar() to print the input characters on the standard output.
• do NOT use any other C library functions. Do your own checking and conversion. However,

  you should NOT use if c==’A’ c==’B’ c==’C’ c==’D’ … c==’Z’. Instead, use the one-statement trick discussed in class. ch as islower(), toupper().

  2.3 Sample Inputs/Outputs (from Standard input):

  red 308 % gcc –Wall –o lab2b lab2B.c red 309 % lab2b
  Hello The World
  HELLO THE WORLD

  How Old Are You?
  

  HOW OLD ARE YOU?
  

  I am 22, and THANKs!
  

  I AM 22, AND THANKS!
  

  ^D (press Ctrl and D)
  

  red 310 %

  2.4 Sample Inputs/Outputs (from redirected input file):

  Using your favorite text editor, create a text file my_input.txt that contains hEllo
  How Are You!
  I Am Good and THAnKs!

  red 311 % lab2b < my_input.txt HELLO
  HOW ARE YOU!
  I AM GOOD AND THANKS!

  red 312 %

  Submit your program by issuing submit 2031Z lab2 lab2B.c
  3. Problem C0 Array and Character array (“Strings”)

  3.1 Specification

  Download program lab2C0.c, compile and run the program.

• Observe that we use marco #define SIZE 20 to define a constant, avoiding magic

  numbers. This is one of the two common ways to define a constant in C. When the program is compiled, the pre-processor replaces every occurrence of SIZE with 20.

• Observe the strange values for elements of array k. Run it again and you might see the change of the strange values. The point here is that, in Ansi-C, an array element is not given an initial value such as 0, rather a garbled value that is randomly generated is given.
  Modify the declaration int k[SIZE]; to int k[SIZE] = {3,5}; Compile and run it again, what do you see? The point here is that if you specify initial values for one or more of the first few elements of the array, the rest elements of the array gets initial value 0.

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Based on this ‘trick’, and without using loop, modify the declaration of k so that k is initialized to all 0s, as shown in the sample output below.

• Observe that for char array msg, which was initilized with a 11 character string literal “Hello World”, the memory size allocated by compiler is 12 bytes. (Observe sizeof operator is used to get the memory size of array.) The extra 1 byte is used to store the null character ‘\0’. On the other hand, a library function call of strlen, which returns the length of string, returns 11. We will discuss string library functions later in class.

  Complete the program by printing out all the elements of array msg, first the encoding (index of the character in the ASCII table), and then the character itself, as shown in the sample output below.
  Observe the special (invisible) character whose index is 0, denoted as ‘\0’, at the end of the array. ‘\0’ is added automatically for you at the end of the array. For more information about “strings”, see section 4.2 below.

• Complete the program by printing out all the elements of array msg2, first the index of the character in the ASCII table, and then the character itself, as shown in the sample output below. Observe that the complier appends ‘\0’ for the rest of space.
  Observe that for msg2 which is declared to be a size 20 array and is initilized with literal “Hello World”, the memory size is 20 bytes, as it declared to be. Despite its larger memory size, however, same as for msg, the library function call of strlen returns 11, and printf also prints Hello World. The point here is that these library functions treat the first ‘\0’ (from left to right) as the end of the string, ignoring values thereafter.
• Remove the comments that around the last 6 lines near the end of the program. Observe that it is msg3, not &msg3, that is passed to scanf as argument. That is, when passing a char array variable to scanf, no & is used. This is a big topic that we will learn later in class. Complete the program by printing out all the elements of array msg3, first the index of the character in ASCII table, and then the character itself, as shown in the sample output below. Complie and run the program, entering a string with no more than 19 characters (why 19?) and without space, such as Helloworld. Observe that the complier may append ‘\0’ for the rest of space. Observe that for msg3 the memory size is 20 bytes, as it declared to be. On the other hand, the library function call of strlen returns 10, and printf prints Helloworld.

  Re-run the program, and enter a string with spaces, e.g., Hello World, and observe that only Hello is stored in the array. Accordingly, printf prints Hello and strlen returns 5.
  This examplifies an issue with reading string simply using scanf(“%s”): it reads input character by character until it encounters a white-space character or a new line character. Thus if the input string contains white spaces it cannot be fully read in. (You are not asked to fix this.)

  Another issue of simply using scanf(“%s”)is that there is no way to detect when the argument array is full. Thus, it may store characters pass the end of the array, causing undfined hehavior (don’t try that). Later we will see other ways to read in a string that contains spaces, and control the input length.

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3.2 Sample Inputs/Outputs

red 361 % gcc lab2C0.c -o lab2c0 red 362 % lab2c0
k[0]: 0
k[1]: 0

k[2]: 0
k[3]: 0
k[4]: 0
k[5]: 0
k[6]: 0
k[7]: 0
k[8]: 0
k[9]: 0

msg: Hello world
memory size of msg: 12 (bytes)
strlen of msg: 11
msg[0] 72 H
msg[1] 101 e
msg[2] 108 l
msg[3] 108 l
msg[4] 111 o
msg[5] 32
msg[6] 119 w
msg[7] 111 o
msg[8] 114 r
msg[9] 108 l
msg[10] 100 d
msg[11] 0

msg2: Hello world
memory size of msg2: 20 (bytes)
strlen of msg2: 11
msg2[0] 72 H
msg2[1] 101 e
msg2[2] 108 l
msg2[3] 108 l
msg2[4] 111 o
msg2[5] 32
msg2[6] 119 w
msg2[7] 111 o
msg2[8] 114 r
msg2[9] 108 l
msg2[10] 100 d
msg2[11] 0
msg2[12] 0
msg2[13] 0
msg2[14] 0
msg2[15] 0
msg2[16] 0
msg2[17] 0
msg2[18] 0
msg2[19] 0

Enter a string: Helloworld msg3: Helloworld
memory size of msg3: 20 (bytes)

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strlen of msg3: 10
msg3[0] H 72
msg3[1] e 101
msg3[2] l 108
msg3[3] l 108
msg3[4] o 111
msg3[5] w 119
msg3[6] o 111
msg3[7] r 114
msg3[8] l 108
msg3[9] d 100
msg3[10]  0
msg3[11]  0
msg3[12]  0
msg3[13]  0
msg3[14]  0
msg3[15]  0
msg3[16]  0
msg3[17]  0
msg3[18]  0
msg3[19]  0

red 342 %

Submit your program using

submit 2031Z lab2 lab2C0.c

You might see other random values from msg3[11]~msg3[19]. It is ok.

4. Problem C Reading and manipulating character arrays

4.1 Specification

Write an ANSI-C program that reads from standard input a word (string with no spaces) followed by a character, and then outputs the word, the number of characters in the word, and the index of the character in the word.

4.2 Useful information

Note that C has no string type, so when you declare literal “hello”, the internal representation is an array of characters, terminated by a special null character ‘\0’, which is added for you automatically. So char helloArr[]= “hello” will give helloArr a representation of
‘h’ ‘e’ ‘l’ ‘l’ ‘o’ ‘\0’ , which has size 6 bytes, not 5.

As shown earlier, one way to read a word from the standard input is to use scanf, which is passed as argument a “string” variable. When using scanf(“%s”, arrayName), the trailing character ‘\0’ is added for you.

4.3 Implementation

• name your program lab2C.c
• define a char array to hold the input word. Assume each input word contains no more than

  20 characters (so what is the minimum capacity the array should be decleared to have?).

• use scanf (“%s %c”, … ) to read the word and char.
• define a function int length(char word[]) which returns the number of

  characters in word (excluding the trailing character ‘\0’). This function is similar to strlen(s) C library function shown earlier, and s.length() method in Java.

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• define a function int indexOf(char word[], char c) which returns the index (position of the first occurrence) of c in word. Return -1 if c does not occur in word. This function is similar to s.indexof()method in Java.
• do not use any existing function from the string library.
• keep on reading until a word “quit” is read in, followed by any character. Hint: You can

  compare the word against the pre-defined terminating token quit by characters. Define a function int isQuit(char word[])which checks if word is “quit”. Later we will learn how to compare two “strings” using library functions). such

  as strcmp).

  4.4 Sample Inputs/Outputs: (output is on a single line)

  red 308 % gcc –Wall lab2C.c –o lab2c
  red 309 % lab2c
  Enter a word and a character separated by blank: hello x
  Input word is “hello”. Contains 5 characters. Index of ‘x’ in it is -1

  Enter a word and a character separated by blank: hello l
  Input word is “hello”. Contains 5 characters. Index of ‘l’ in it is 2

  Enter a word and a character separated by blank: beautifulWord u
  Input word is “beautifulWord”. Contains 13 characters. Index of ‘u’ in it is 3

  Enter a word and a character separated by blank: quit x red 310 %

  Submit your program by issuing submit 2031Z lab2 lab2C.c 5. Problem D Array, Character array / String

  5.1 Specification

  Write an ANSI-C program that takes input from Standard in, and outputs the occurrence count of digits 0-9 in the inputs.

  5.2 Implementation

• name your program lab2D.c
• use getchar and putchar to read from the standard input. The program continues to

  read until EOF is entered. Then outputs the occurrence count of each digit in the input.

• use an integer array as the counters, don’t use 10 individual counters.
• when a character is read in, how to find the corresponding counter? Don’t use statement

  such as if (c==’0′) … else if (c==’1′) … else if (c==’2′)… else… Instead, use the one-statement trick discussed in class, which takes advantage of the index of the character to figure out the corresponding counter in the array.

  5.3 Sample Inputs/Outputs: (download the input file input.txt)

  red 368 % gcc -Wall lab2D.c -o lab2d red 369 % lab2d
  YorkU LAS C
  ^D (press Ctrl and D)

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0: 0
1: 0
2: 0
3: 0
4: 0
5: 0
6: 0
7: 0
8: 0
9: 0
red 370 % lab2d EECS2031Z CB121
^D (press Ctrl and D) 0: 1

1: 3
2: 2
3: 1
4: 0
5: 0
6: 0
7: 0
8: 0
9: 0
red 371 % lab2d EECS3421 thiw is good 3 address 500 yu266074 423Dk

^D (press Ctrl and D)

0: 3
1: 1
2: 3
3: 3
4: 3
5: 1
6: 2
7: 1
8: 0
9: 0
red 372 % lab2d < input.txt 0: 4

1: 7
2: 4
3: 4
4: 5
5: 2
6: 2
7: 5
8: 0
9: 0
red 373 %

Submit your program by issuing

submit 2031Z lab2 lab2D.c

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6. Problem E0 ‘Boolean’ in 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. Download program Lab2E0.c, compile and run it.
• 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
o && return 1 if both operands are non-zero, return 0 otherwise. || return 1 if either

operand is non-zero, and return 0 otherwise.

• Assume the author mistakenly use = , rather than ==, in the three if conditions. Observe that although x has initial value 100, both if(x=3) and if(x=4) clauses were executed. This illurstrates a few interesting things in Ansi-C:

o Unlike a Java complier, gcc does not treat this as a systax error.
o Assignment experssion such as x=3 has a return value, which is the value being

assigned to. So if(x=3) becomes if(3), and if(x=0) becomes if(0)
o Anynon-zeronumberistreatedas‘true’inselectionstatement.Thusif(x=3)and

if(x=-4) 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.

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 in the sample

outputs below.

      hello 0
      hello 1
      hello 2
      hello 3
      hello 4
      hello 5
      hello 6
      hello 7
      hello 8

No submissions for problem E0.

7. Problem E scanf, arimthemic and logicl operators

7.1 Specification

Write an ANSI-C program that reads input from Standard in, which represents a year, and then determines if the year is a leap year.

7.2 Implementation

• name your program lab2E.c

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• keep on readin a (4 digit) integer of year, until a negative number is entered.
• put the definition (implementation) of function isLeap after your main function.
• display the prompt Enter a year:
• for each non-negative input, display the outputs Year x is a leap year
  or
  Year x is not a leap year

  7.3 Sample Inputs/Outputs:

  red 364 % gcc -Wall lab2E.c -o lab2E red 365 % lab2E
  Enter a year: 2010
  Year 2010 is not a leap year

  Enter a year: 2012
  Year 2012 is a leap year

  Enter a year: 2017
  Year 2017 is not a leap year

  Enter a year: 2018
  Year 2018 is not a leap year

  Enter a year: 2019
  Year 2019 is not a leap year

  Enter a year: 2020
  Year 2020 is a leap year

  Enter a year: 2200
  Year 2200 is not a leap year

  Enter a year: 2300
  Year 2300 is not a leap year

  Enter a year: 2400
  Year 2400 is a leap year

  Enter a year: 2500
  Year 2500 is not a leap year

  Enter a year: -6 red 366 %

  Submit your program by issuing

                      submit 2031Z lab2 lab2E.c
  

• define a ‘Boolean’ function 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.

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In summary, for lab2 you should sbmit the following files:

lab2A.c lab2B.c lab2C0.c lab2C.c lab2D.c lab2E.c

Common Notes

All submitted files should contain the following header:

/*************************************** * EECS2031 – Lab2 *
* Author: Last name, first name *
* Email: Your email address *

* EECS_num: Your EECS login username *
* York Student #: Your student number * ****************************************/

In addition, all programs should follow the following guidelines:

• Include the stdio.h library in the header of your .c files.
• Use /* */ to comment your program. You are not encouraged to use //.
• Assume that all inputs are valid (no error checking is required, unless

  asked to do so).

  You are also encouraged to

• Give a return type int for main(), and return 0 at the end of main()
• Specify parameters for main() function, as main(int argc, char *argv[])