Need help getting program print out the digit in words

I need to have the out come of
Enter a digit between 0 and 9: 4
You entered the number four

So far I have come up with this

insert [/

Code:

---

*
File: wa2.c
Description: Written Assignment #2 for COS-116. Have the program print out the digit in words
Programmer: Thomas Garnick
Date: April 23rd 2008
*/
#include <stdio.h>
int main()
{
    char one, two, three, four, five;
    int number;

printf("Input Number: ");
scanf("%d",&number);
if (number == '1')
number = one;
if (number == '2')
number = two;
else if (number == '3')
number = three;
else if (number == '4')
number = four;
else if (number == '5')
number = five;

printf("you entered the number %c", number);

return 0;
}

---

]

When I enter a number it gives me a special character as my output. We are still in the basics and have not gotten into character strings. Any help would be greatly apritiated

If you haven't gotten into character strings you can't do the problem.

I'm assuming you have gotten into string literals; you should note that "one" and one are not the same thing, and that "one" cannot be fit into a char variable.

for any one that is currious I finally solved it.

Code:

---

/*
File: wa2.c
Description: Written Assignment #2 for COS-116. Dispays #'s 1-9 as a word
Programmer: Thomas Garnick
Date: April 23rd 2008
*/
#include <stdio.h>
int main()
{
    int number;

printf("Input a number 1-9: ");
scanf("%d", &number);
if (number == 0)
printf("\nYou entered the number Zero\n");
else if (number ==1)
printf("\nYou entered the number One\n");
else if (number == 2)
printf("\nYou entered the number Two\n");
else if (number == 3)
printf("\nYou entered the number Three\n");
else if (number == 4)
printf("\nYou entered the number Four\n");
else if (number == 5)
printf("\nYou entered the number Five\n");
else if (number == 6)
printf("\nYou entered the number Six\n");
else if (number == 7)
printf("\nYou entered the number Seven\n");
else if (number == 8)
printf("\nYou entered the number Eight\n");
else if (number == 9)
printf("\nYou entered the number Nine\n");
else
printf("\nSorry invalid number\n"
                "\nSince I'm a simple program I wont be able to tell you the answer\n"
                "\nThanks for trying\n");

return 0;
}

---

I think switch statement will look better here

Yet another possibility is to use an array of strings. If the number is in the range [0,9], display the corresponding array element, otherwise display the error message.

laserlight has an excellent idea. The code might look something like this. Just to clarify.

Code:

---

#include <stdio.h>


int main(void)
{
        int number=0 ;
        char numbers[10][6] = { "Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" };

        do {
                printf("Input a number [0-9]: ");
                scanf("&#37;d", &number);

                if ( !(number >= 0 && number <=9) )
                        printf("Sorry input outside of valid range: Try again Please\n");

        } while ( !(number >= 0 && number <=9) );

        printf("You entered the number %s\n", numbers[number] );

return 0;
}

---

Actually, I would rather create numbers as:

Code:

---

const char *numbers[] = {"Zero", "One", "Two", "Three", "Four",
                        "Five", "Six", "Seven", "Eight", "Nine"};

---

since it is an array of pointers to the first characters of ten string literals.

cute, ok. I agree totally... plus some error checking ;-)

Code:

---

#include <stdio.h>

int main()
{
        int number=10;
        const char *numbers[] = { "Zero", "One", "Two",  "Three", "Four",
                                  "Five", "Six", "Seven", "Eight", "Nine"  };

        printf("Input a number [0-9]: ");

        while ( !scanf("&#37;d", &number) || !(number >= 0 && number <=9) )
        {
                printf("Sorry input outside of valid range: Try again Please\n");
                printf("Input a number [0-9]: ");
                fflush(stdin);
        }
        printf("You entered the number %s\n", numbers[number] );

return 0;
}

---

Code:

---


#include<stdio.h>
int main()
{
 
    int choice;
    printf("enter the any number between 0 and 9\n");
    scanf("%d", &choice);
    switch(case)
  {
    case 1: printf("ONE");
                break;
    case 2: printf("TWO");
                break;
    case 3: printf("THREE");
                break;
    case 4: printf("FOUR");
                break;
    case 5: printf("FIVE");
                break;
    case 6: printf("SIX");
                break;
    case 7: printf("SEVEN");
                break;
    case 8: printf("EIGHT");
                break;
    case 9: printf("NINE");
                break;
    defualt: printf("You are entered wrong number\n");
}

---

Quote:

---

Originally Posted by sillyman

cute, ok. I agree totally... plus some error checking ;-)

---

You should review your error checking

Quote:

---

The return value of scanf() is the number of variables that were successfully assigned values, or EOF if there is an error.

---

So your code perfoms the same way when scanf returns 1 or EOF. I doubt you want it that way

Why would a college student working on an exercise type in EOF on a simple programming example? Can you give me an example of how to enter EOF into the program?

Don't flush stdin!
http://cpwiki.sourceforge.net/Common...kes_and_errors

Quote:

---

Originally Posted by sillyman

Why would a college student working on an exercise type in EOF on a simple programming example? Can you give me an example of how to enter EOF into the program?

---

You don't enter EOF. EOF is returned if there was an error. For example, if the user entered "Hi there!" instead of 1.

Quote:

---

Originally Posted by Elysia

Don't flush stdin!

---

Okay, thanks. I never tried it before. So a better option would be to use non-buffered input?

If you need to clear the input buffer, there's an alternative way. Check the link. It contains links to all you need to know.

Quote:

---

Originally Posted by Elysia

Don't flush stdin!
http://cpwiki.sourceforge.net/Common...kes_and_errors


You don't enter EOF. EOF is returned if there was an error. For example, if the user entered "Hi there!" instead of 1.

---

Actually, according to the specs, EOF is returned "If the first attempt to read a character returns EOF" (or end of string in the case of sscanf). This means that EOF will only happen if:
1. input is coming from a text-file (and the program tries to read something when there's nothing left in the file).
2. the user types CTRL-Z (or CTRL-D in a Unix/Linux environment).

If nothing could be read in, then it will return 0. For example when a number is requested and "blah" is given. [That is for decimal numbers, hex numbers would accept "b" as input, and lah would be left in the buffer].

--
Mats