Homework help

This is a discussion on Homework help within the C Programming forums, part of the General Programming Boards category; Originally Posted by @nthony requesting obligatory posting of mac's signature and lock... in that order. You rang, sir?...

  1. #16
    Deathray Engineer MacGyver's Avatar
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    Quote Originally Posted by @nthony View Post
    requesting obligatory posting of mac's signature and lock... in that order.
    You rang, sir?

  2. #17
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    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?

  3. #18
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    Smile I dont really understand subquestion L

    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;
    }

  4. #19
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    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.

  5. #20
    CSharpener vart's Avatar
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    BLUR = BLUR[0]
    not exactly

    Code:
     BLUR == &BLUR[0]
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  6. #21
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    Quote Originally Posted by mkdl750 View Post
    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;
    }

  7. #22
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    Okay now please show the memory diagram in subquestion L.

  8. #23
    Woof, woof! zacs7's Avatar
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    > Okay now please show the memory diagram in subquestion L.
    Negative. http://pages.cs.wisc.edu/~cs302/reso..._diagrams.html

  9. #24
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    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?

  10. #25
    Deathray Engineer MacGyver's Avatar
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    That code is horrible. If someone taught you to write that, they did you a disservice.....

  11. #26
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    any further assistance about my answer to subquestion E?

  12. #27
    and the Hat of Guessing tabstop's Avatar
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    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.

  13. #28
    Malum in se abachler's Avatar
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    Code:
     
    char Dec(char a){
       return 64 + (((63 & a)+ 1) &#37; 26);
       }
    do I win a cookie ?
    Last edited by abachler; 07-11-2008 at 10:56 AM.
    Until you can build a working general purpose reprogrammable computer out of basic components from radio shack, you are not fit to call yourself a programmer in my presence. This is cwhizard, signing off.

  14. #29
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    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);
    }

  15. #30
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    Quote Originally Posted by mkdl750 View Post
    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;   
    }

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