Homework help

I am a 16-year old Filipino who is a 2nd year college student at Informatics Computer Institute (Malolos, Bulacan branch). I have a coursework assignment in CS215 (C Programming) and I need your extensive help. Please post the following codes so that I only need to worry about the compilation.

Here are the requirements for the assignment:

1. Do not add any nonsense in my C code, for my lecturer will not accept any nonsense added in my code!!!
2. Be able to demonstrate and explain programs to your lecturer.
3. Submit a physical file holding the following:
a) Printed hard copy of all programs
b) Printed test cases
c) All soft copy programs in a CD-R (please attach firmly to
physical file)
4. Do what is exactly asked in the questions.

The assignment is worth 40 points.
The deadline is July 25, 2008, so you better hurry up.

Question 1 (24 points)

Question 1 contains 12 sub-questions which will be in separate C programs. I'll be the one who will design the filenames for each program.

Each of the functions below needs to be run in a program. Each of the programs should have at least one test case (with a screen capture) which shows the succesful running of the program. Each sub-question is worth 2 points.

a) Write a function called ChangeToNext that takes in an uppercase character parameter ch. The function returns the next uppercase character if ch is between 'A' and 'Y'. If ch == 'Z', the function returns 'A'. You may assume that the character parameter ch is always an uppercase alphabetic character.

b) Write a function called Difference that takes in an integer array intarr of size 10. The function returns the difference between the sum of all values from cell 0 to 4 and the sum from cell 5 to 9. For example, if intarra[10]={1,1,1,1,1,2,2,2,2,2}, Difference (intarr) returns -5 ((1+1+1+1+1) - (2+2+2+2+2)

c) Write a function PerimeterMinusCenter that takes in a two-dimensional array called intarr of 3 rows and 3 columns. The function sums the values in the perimeter of the array. It then returns the difference of this sum and the center value (the value in cell row 1, column 1 or [1][1]).

For example, if

Code:

---

intarr[3][3] = {1,1,1}
                                {1,2,1}
                                {1,1,1}
                                };

---

PerimeterMinusCenter(intarr) returns 6 (1+1+1+1+1+1+1+1 - 2).

d) Write a function Compare that takes in a two-dimensional array called intarr of 3 rows and 3 columns. The function sums up separately the values in the diagonal of the array and the non-diagonal values. It then compares these two sums and return 1 if the diagonal sum is larger than the non-diagonal sum or 0 otherwise.

For example, if

Code:

---

intarr[3][3] = {{1,2,2}
                                  {2,1,2}
                                  {2,2,1}
                                  };

---

Compare(intarr) returns 0 (since (1+1+1) < (2+2+2+2+2+2)).

e) Write a recursive function called CountEvenASCII that takes in a string parameter str and return the number of characters with even ASCII codes. For example CountEvenASCII("ABCDE") returns 2 (as B and D are of even ASCII codes).

f) Change your solution above (sub-question e) into an iterative solution.

g) Write a function called ReturnNextCase that takes in a character parameter ch and returns the next alphabet in the other case. For example, ChangeCase('A') returns 'b' and ChangeCase('a') returns 'B'. You may always assume the character parameter ch is always alphabetic.

h) Write a function called ReturnASCIIDifference that takes in a string parameter str. The function sums up separately even ASCII characters and odd ASCII characters and return their difference. For example, ReturnASCIIDifference('ABCD') returns 2 (as the ASCII values of 'A', 'B', 'C', and 'D' are 65, 66, 67, and 68 respectively and (66+68) - (65+67) = 2).

i) Write a function called CheckPosNeg that takes in a reference integer parameter num. The function returns 0 if the content of the reference parameter is a positive number and 1 otherwise.

j) Write a function called PrintToZero that takes in a reference integer parameter num. The function prints out the content of num up to zero. For example, if *num=10, then PrintToZero(num) prints 10 9 8 7 ... 0..

lol, read the rules.

Sorry, sir, I'm just a noob and I need you help. Please post the codes if necessary.

But it's your assignment? Read http://cboard.cprogramming.com/annou...t.php?f=4&a=39

What happens when you finish your course (if you do...)? They're testing you, not us. If you have a specific question, feel free to ask. ie, part 'a':

Code:

---

char ChangeToNext(char ch)
{
    /* insert code here */
}

---

... and what's to say we don't contact the institute and tell them you're cheating? It's only fair. Not to mention 3 weeks is plenty of time.

If you put as much effort into your programming as you do with your presentation and sucking up skills, then you might actually get somewhere.

http://www.catb.org/~esr/faqs/smart-....html#homework

Okay sorry about that I'll be the one to design the codes myself and need further help later on.

requesting obligatory posting of mac's signature and lock... in that order.

Code:

---

#include <stdio.h>
char ChangeToNext(char ch);
main()
{
    char x;
    printf("Enter uppercase letter: ");
    fflush(stdin);
    scanf("&c",&x);
    if (x == 'A' || x == 'B' || x == 'C' || x == 'D' || x == 'E' || x == 'F' ||
    x == 'G' || x == 'H' || x == 'I' || x == 'J' || x == 'K' || x == 'L' ||
    x == 'M' || x == 'N' || x == 'O' || x == 'P' || x == 'Q' || x == 'R' ||
    x == 'S' || x == 'T' || x == 'U' || x == 'V' || x == 'W' || x == 'X' ||
    x == 'Y' || x == 'Z')
    {
        ChangeToNext(x);
    }
    else
    {
        printf("\nWrong character. Try again.\n");
    }
    getch();
    return 0;
}
char ChangeToNext(char ch)
{
    printf("\nThe next alphabet letter is: ");
    if (ch == 'A')
    {
        printf("B");
    }
    else if (ch == 'B')
    {
        printf("C");
    }
    else if (ch == 'C')
    {
        printf("D");
    }
    else if (ch == 'D')
    {
        printf("E");
    }
    else if (ch == 'E')
    {
        printf("F");
    }
    else if (ch == 'F')
    {
        printf("G");
    }
    else if (ch == 'G')
    {
        printf("H");
    }
    else if (ch == 'H')
    {
        printf("I");
    }
    else if (ch == 'I')
    {
        printf("J");
    }
    else if (ch == 'J')
    {
        printf("K");
    }
    else if (ch == 'K')
    {
        printf("L");
    }
    else if (ch == 'L')
    {
        printf("M");
    }
    else if (ch == 'M')
    {
        printf("N");
    }
    else if (ch == 'N')
    {
        printf("O");
    }
    else if (ch == 'O')
    {
        printf("P");
    }
    else if (ch == 'P')
    {
        printf("Q");
    }
    else if (ch == 'Q')
    {
        printf("R");
    }
    else if (ch == 'R')
    {
        printf("S");
    }
    else if (ch == 'S')
    {
        printf("T");
    }
    else if (ch == 'T')
    {
        printf("U");
    }
    else if (ch == 'U')
    {
        printf("V");
    }
    else if (ch == 'V')
    {
        printf("W");
    }
    else if (ch == 'W')
    {
        printf("X");
    }
    else if (ch == 'X')
    {
        printf("Y");
    }
    else if (ch == 'Y')
    {
        printf("Z");
    }
    else if (ch == 'Z')
    {
        printf("A");
    }
    else
    {
        printf("\nError.");
    }
    return;
}

---

Here's sub-question A. I have to display the next uppercase letter of the alphabet (in this case 'Z' has to roll back to 'A') and everytime i finish my input, I get a "Wrong character" message. Any further tweaks on this one?

I almost forgot the "scanf("&c",&x);" in line 8 is actually "scanf("&#37;c",&x);". Thank you for your consideration in subquestion A. I'm advancing to subquestion B.

I have to write a function called "Difference" that takes in an integer array "intarr" of size 10. The function returns the difference between the sum of all values from cell 0 to 4 and the sum from cell 5 to 9. For example, if intarra[10]={1,1,1,1,1,2,2,2,2,2}, Difference(intarr) returns -5 ((1+1+1+1+1) - (2+2+2+2+2). Any valuable hints?

Here's the start of my code:

Code:

---


#include <stdio.h>
void Difference[int intarr[10]
main()
{ //code goes here
}

---

Quote:

---

Any valuable hints?

---

Yes. You need to read your textbook and/or notes again concerning function syntax. For example,

Code:

---

void Difference[int intarr[10]

---

should be the function prototype:

Code:

---

void Difference(int intarr[10]);

---

The "10" is optional in this case, since it will be ignored by the compiler. But note that you declared the function's return type as void, but your example has the function returning -5. Clearly, there is an inconsistency there.

Also, the global main function returns an int, so you should explicitly define it as such. You should also explicitly return 0 at the end of the global main function, though this is optional with respect to the 1999 edition of the C standard.

Code:

---


#include "stdio.h"
char ChangeToNext(char ch);
int main(int argc, char* argv[])
{

  char ch;
  char Next_ch;
  printf("enter the character\n");
  scanf("%c", &ch);
  Next_ch= ChangeToNext(ch);
  printf("%c", ch);
        return 0;
}
char ChangeToNext(char ch)
{
  if(ch>'Y'&&ch<'N')
  {
    return ++ch;
    }
    else if (ch=='N')
    {
      return 'Y';
    }
}

---

shwetha_siddu, do not post homework solutions unless it is clear that the member has sufficiently attempted to solve the question. To this end, I have modified your code example so that it will not produce the correct answer.

EDIT:
Actually, come to think of it, your code is incorrect to begin with since your ChangeToNext() implementation has one control path that does not return a value. Oh well.

> mkdl, this is a really not a good way to do this.
What I do suggest is to use char() - e.g: char(65), output: A
Something like this:

Code:

---

char GetNext(char a)
{
 if(a == 'Z')
    return 'A';
 else
 {
  int dec = (int)a + 1;
  return char(dec);
  }
}

---

About the other assignments; these things are really basic and shouldn't take long, just go over what you were taught in class (strings library functions, casting, loops etc')

Well, I was kinda bored so...
e)

Code:

---

int rec(char *str)
{
 if(str[0] == '\0')
        return 0;
 if((int)str[0] % 2 == 0)
        return 1 + rec(str + 1);
 else
        return rec(str + 1);
}

---

The iterative solution is very basic, if you are still having troubles, look for strlen.


EDIT: If I'm overriding a rule, please edit my post, thanks.

Quote:

---

Originally Posted by mkdl750

Sorry, sir, I'm just a noob and I need you help. Please post the codes if necessary.

---

http://www.ctrlaltdel-online.com/comic.php?d=20060823
...cause everything is about definitions....:)

Quote:

---

Originally Posted by @nthony

requesting obligatory posting of mac's signature and lock... in that order.

---

You rang, sir? :)

2. Be able to demonstrate and explain programs to your lecturer.

Even if we create the code for you, do you think you can able to explain it?

Okay I'm done with subquestions A, B, C, D, and I really don't understand what memory diagrams are. I am asked to trace the following C program and write down its output. Can you kindly assist me in this one?

Code:

---

#include <stdio.h>
#include <conio.h>

char *B1, *B2, *B3, *B4, *B5;
char BLUR[8] = {'M', 'O', 'R', 'E', 'B', 'L', 'U', 'R'};

main()
{
clrscr();
B5=B4=BLUR;
++B4;
B3=&BLUR[6];
B2=B5+2;
B1=&BLUR[7]-3;

printf ("%c\t%c\t%c\t%c\t%c",*B1,*B2,*B3,*B4,*B5);

getch();
return 0;
}

---

B5=B4=BLUR


1) B5 and B4 are pointers to characters

2) BLUR = BLUR[0]

I think that above two points will help you to calcuate the output.

Quote:

---

BLUR = BLUR[0]

---

not exactly

Code:

---

BLUR == &BLUR[0]

---

Quote:

---

Originally Posted by mkdl750

Okay I'm done with subquestions A, B, C, D, and I really don't understand what memory diagrams are. I am asked to trace the following C program and write down its output. Can you kindly assist me in this one?

Code:

---

#include <stdio.h>
#include <conio.h>

char *B1, *B2, *B3, *B4, *B5;
char BLUR[8] = {'M', 'O', 'R', 'E', 'B', 'L', 'U', 'R'};

main()
{
clrscr();
B5=B4=BLUR;
++B4;
B3=&BLUR[6];
B2=B5+2;
B1=&BLUR[7]-3;

printf ("%c\t%c\t%c\t%c\t%c",*B1,*B2,*B3,*B4,*B5);

getch();
return 0;
}

---

---

The output of following program is :- B R U O M

Code:

---


#include <stdio.h>
#include <conio.h>
char *B1, *B2, *B3, *B4, *B5;
char BLUR[8] = {'M', 'O', 'R', 'E', 'B', 'L', 'U', 'R'};

int main()
{
B5=B4=BLUR;//Both Pointers are assigned to 'M'
++B4;  //Now it will point to 'O'
B3=&BLUR[6]; //'U' has been assigned
B2=B5+2; //'R' has been assigned
B1=&BLUR[7]-3; //'B' has been assigned

printf ("%c\t%c\t%c\t%c\t%c",*B1,*B2,*B3,*B4,*B5); //B  R  U  O  M

getch();
return 0;
}

---

Okay now please show the memory diagram in subquestion L.

> Okay now please show the memory diagram in subquestion L.
Negative. http://pages.cs.wisc.edu/~cs302/reso..._diagrams.html

Heres my draft solution for subquestion E.

Code:

---

#include <stdio.h>
#include <string.h>

int CountEvenASCII(char *str);

main()
{
char *x[20];
printf("Enter a string of letters: ");
scanf("%c",x);
CountEvenASCII(*x);
printf("%d",CountEvenASCII(*x));
getch();
return 0;
}

int CountEvenASCII(char *str)
{
 if(str[0] == '\0')
        return 0;
 if((int)str[0] % 2 == 0)
        return 1 + CountEvenASCII(str + 1);
 else
        return CountEvenASCII(str + 1);
}

---

But how can I display the number of letters with even ASCII character values?

That code is horrible. If someone taught you to write that, they did you a disservice.....

any further assistance about my answer to subquestion E?

1. There is no such thing as "main()". There is "int main()", and there is "int main(int argc, char *argv[])".
2. Indentation is something you should look into. Basically: everything inside curly braces should be moved over by the same amount (and the more braces, the further over you go).
3. A string (which should probably not be called "x") can be either char *x, or char x[20], but not both. The declaration char *x[20] gives you 20 pointers-to-characters; each pointer-to-character could be a string one day, if you malloc'ed memory for those pointers to point to.
4. %c only reads one character from standard input. %s will read in a string, provided it has no spaces, but needs work to be safe. There is such a thing as fgets.
5. There's no need to call the function twice -- once inside the print statement should be sufficient.

Code:

---


char Dec(char a){
  return 64 + (((63 & a)+ 1) % 26);
  }

---

do I win a cookie ?

Can you help me correct this code?

Code:

---


#include <stdio.h>
#include <string.h>

int CountEvenASCII(char *str);

main()
{
char *x[20];
printf("Enter a string of letters: ");
scanf("&#37;c",x);
CountEvenASCII(*x);
printf("%d",CountEvenASCII(*x));
getch();
return 0;
}

int CountEvenASCII(char *str)
{
 if(str[0] == '\0')
        return 0;
 if((int)str[0] % 2 == 0)
        return 1 + CountEvenASCII(str + 1);
 else
        return CountEvenASCII(str + 1);
}

---

Quote:

---

Originally Posted by mkdl750

Can you help me correct this code?

Code:

---


#include <stdio.h>
#include <string.h>

int CountEvenASCII(char *str);

main()
{
char *x[20];
printf("Enter a string of letters: ");
scanf("&#37;c",x);
CountEvenASCII(*x);
printf("%d",CountEvenASCII(*x));
getch();
return 0;
}

int CountEvenASCII(char *str)
{
 if(str[0] == '\0')
        return 0;
 if((int)str[0] % 2 == 0)
        return 1 + CountEvenASCII(str + 1);
 else
        return CountEvenASCII(str + 1);
}

---

---

Your algorithm is correct but the way you are calling it is wrong.
This should be done just like the following (check out the changes in CountEvenASCII)

Code:

---

int CountEvenASCII(const char *str) // since you are not making any changes to the string, it's supposed to be a constant string.
{
 if(str[0] == '\0')
        return 0;
 if((int)str[0] % 2 == 0)
        return 1 + CountEvenASCII(str + 1);
 return CountEvenASCII(str + 1); // no 'else' is needed.
}

int main(void)
{
    const char *abc = "AB";
    printf("%d", CountEvenASCII(abc));
    return 0; 
}

---

Okay, how can I tweak the input sequence of subquestion E so that I can be able to finish the program?

My first solution for subquestion J:

Code:

---


#include <stdio.h>

int PrintToZero (int *num, int counter);

main()
{
    int *x, y;
    x = &y;
    printf("Input number: ");
    scanf("%d",&x);
    printf("\n");
    PrintToZero(x,y);
    getch();
    return 0;
}

int PrintToZero (int *num, int counter)
{
    for (counter=0;counter<*num;counter++)
    {
        printf("%d ",*num);
        *num = *num - 1;
    }
    return;
}

---

Theres some mistake here, but it should print like "n n-1 n-2..... 0" (if n = 10, it would read "10 9 8 7 6 5 4 3 2 1 0")

The problem is in the scanf statement. Why try to read an integer into a variable that is not of type int?

what is an iterative statement?

Quote:

---

Originally Posted by mkdl750

what is an iterative statement?

---

www.dictionary.com

You should have posted the meaning here. What I'm talking about is the iterative function in C.

Quote:

---

Originally Posted by mkdl750

You should have posted the meaning here. What I'm talking about is the iterative function in C.

---

You say that as though it exists.

There are functions that are iterative in nature (the opposite of iterative is recursive). But iterative is not a keyword in C, nor is it language-specific--we are using the word in its ordinary English usage.

Code:

---

#include <stdio.h>
#include <string.h>
void CountEvenASCII(char str[5]);

main()
{
    char w[5];
    printf("Input character string: ");
    scanf("&#37;s",w);
    CountEvenASCII(w);
    getch();
    return 0;
}

void CountEvenASCII(char str[5])
{
    int x,y,z,c,e;
    e=0;
    x=strlen(str);
    x=x-1;
    for (y=0;y<x;y++);
    {
        z=x-y;
        c=str[z];
        if (c % 2 == 0)
        {
            e=e+1;
        }
    }
    printf("\nNumber of characters with even ASCII numerical values: %d\n\n",e);
}

---

This code above should display the number of letters with even ASCII values, but it doesn't work (It just displays if the 1st character has an even ASCII value). Any further assistance or tweaking please?

Please read the note inside the program.

Code:

---


//display number of letters with even ascii values - help
#include <stdio.h>
#include <string.h>
void CountEvenASCII(char str[5]);

main()
{
    char w[5];
    printf("Input character string: ");
    scanf("&#37;s", w);
    CountEvenASCII(w);
    getch();
    return 0;
}

void CountEvenASCII(char str[5])
{
  int c, even, i;
  even = i = 0;
  while(str[i] != '\0')  {
      if(str[i] % 2 == 0)
          even++;
      i++;
  }


/*  Please don't use variables of x,y,z,c,e. Aside from the commonly used
    loop counters (i, j, k, m, etc), variables with meaningless names just
    confuse everyone, including yourself. 
    int x,y,z,c,e;
    e=0;
    x=strlen(str);
    x=x-1;
    for (y=0;y<x;y++);
    {
        z=x-y;
        c=str[z];
        if (c % 2 == 0)
        {
            e=e+1;
        }
    }
*/
    printf("\nNumber of characters with even ASCII numerical values: %d\n\n", even);

}

---

Quote:

---

Originally Posted by mkdl750

Code:

---

#include <stdio.h>
#include <string.h>
void CountEvenASCII(char str[5]);

main()
{
    char w[5];
    printf("Input character string: ");
    scanf("%s",w);
    CountEvenASCII(w);
    getch();
    return 0;
}

void CountEvenASCII(char str[5])
{
    int x,y,c,e;
    e=0;
    x=strlen(str);
    x=x-1;
    for (y=0;y<x;y++);
    {
        c=str[y];
        if (c % 2 == 0)
        {
            e=e+1;
        }
    }
    printf("\nNumber of characters with even ASCII numerical values: %d\n\n",e);
}

---

This code above should display the number of letters with even ASCII values, but it doesn't work (It just displays if the 1st character has an even ASCII value). Any further assistance or tweaking please?

---

Should be fixed now. I'm not sure what you were trying to accomplish with the z variable, but it wasn't helping.

Also, you set a string length of 5, meaning if you type a sentence of more than 4 characters you're going to get a buffer overrun error. Better expand that. I'd say 255.

Hmmm, spaces are #32 on the ascii table. I wonder about punctuation marks too. Will you need to take those into account? In all honesty, as beginner as you are I'm sure your teacher isn't going to dock you for it too badly.

Lemme give you one more hint. e++; means the exact same thing as e=e+1; and x--; means the exact same thing as x=x-1;. They're called increment or decrement operators and are a sort of short hand for this exact sort of thing which turns up a ton.

thanks for your solution :D

Okay how I change the recursive solution into iterative ones? just remove the while?

how do I convert subquestion e into an iterative statement?

There are no recursive solutions shown on this page. They're all iterative one's.

Removing the while will not make an iterative loop into a recursive solution. You need to google and read up on the terms "iterative" and "recursive".

Code:

---

#include <stdio.h>
#include <string.h>
void CountEvenASCII(char str[5]);

main()
{
    char w[5];
    printf("Input character string: ");
    scanf("&#37;s", w);
    CountEvenASCII(w);
    getch();
    return 0;
}

void CountEvenASCII(char str[5])
{
  int c, even, i;
  even = i = 0;
  while(str[i] != '\0')  {
      if(str[i] % 2 == 0)
          even++;
      i++;
  }

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If Mr. Adak's solution above is an iterative statement, how do I turn it into a recursive one?

Okay, first things first. Do you understand how the solution above works? Judging by the code you wrote I'd guess you have a 50/50 chance of. So step one, explain to me in words why his solution works, what it does.