Homework help

**mkdl750** · 07-13-2008

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

**mkdl750** · 07-15-2008

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")

**tabstop** · 07-15-2008

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

**mkdl750** · 07-15-2008

what is an iterative statement?

**tabstop** · 07-15-2008

Originally Posted by mkdl750

what is an iterative statement?

www.dictionary.com

**mkdl750** · 07-15-2008

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

**tabstop** · 07-15-2008

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.

**mkdl750** · 07-17-2008

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?

**Adak** · 07-17-2008

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);

}

**guesst** · 07-17-2008

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.

**mkdl750** · 07-17-2008

thanks for your solution

**mkdl750** · 07-17-2008

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

**mkdl750** · 07-17-2008

how do I convert subquestion e into an iterative statement?

**Adak** · 07-17-2008

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".

**mkdl750** · 07-17-2008

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++; 
   }

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