Please Help A Noob Here (plz Im Begging You)

hi all
im really new to c programming so i just need some help from you guys to get me started
i got to do this task to create a program that will tell the user whether it is a leap yr and how many days there are in the month when enter...
so far i got this:

printf("Enter yr and month:");
scanf("%d", &a);
if *i got this type just working on it so ignore me*

{
printf("has (366) days\n");}
else{
printf("has (365) days\n");}

scanf("%d", &month);
if (month=1);
{
printf("Janurary has 31 days\n");}

else if (month=2);
{
printf("Feburary has 28 days\n");}
*/
}
return 0;
}



what did i do wrong???


and later on when im required to change the year like to proper words... how do i do it?

EG:

Enter year and month: 1980 2
nineteen hundred eighty<--------------- how do i do this for years from 1792 - 9999???
has three hundred and sixty six (366) days
February has 29 days

THANK YOU ALL FOR YOUR HELP AND I REALLY APPRECIATE IT

Code:

---

if (month=1);

---

A) Doesn't do what you want. It does:

Code:

---

if ((month = 1) != 0)

---

Since 1 is always not equal to zero, what you've achieved is

Code:

---

month =1; if (1) ...

---

B) Contains only an empty statement ";" - so nothing actually gets done in the if-statement.

This in itself will probably lead to the "else if" part further down being an error, because the else doesn't belong to an if-statement.

--
Mats

first of all thx a lot for ur reply
im only been learning c for a couple of days and my teacher is being a jackass
so how do i fix the if argument if i want my program to acknowledge that the 1 being entered is actually a month? how do i creat a printf statement for this if statemnet

You probably want

Code:

---

if(month == 1) ...

---

There should never be a semicolon DIRECTLY after the parenthesis of an if-statement, as that means that the if-statement contains no code to execute for the "true" choice, which means that there's no need to have an if in the first place. [Unless you have an else after it, in which case the condition should be the opposite to what it is right now].

By the way, for days per month, a switch-statement is probably more suitable:

Code:

---

switch(month)
{
  case 4:
  case 6:
  case 9:
  case 11:
      ... 30 days.
      break;
  case 2:
      ... 28 or 29  days.
      break;
  default:
      if (month < 1 || month > 12)  ... illegal month. ;
      else ... 31 days
      break;
}

---

Only use this if your teacher has introduced "switch" however.


--
Mats

what other options are there if the bastard hasnt taught switch yet? thx a lot for ur time

Code:

---

if(month == x) ...
else if (month == y) ...
...
...

---

Of course, you only need to check some of them, like in the switch. If you KNOW that the value is between 1 and 12, then there are 5 months that aren't 31 days: February, April, June, September and October, so those are the only ones you need an if-statement for, the rest are 31 days because that's the only "option left".

--
Mats

thx for the reply
how do i line up my if statements and else statements if there are several of them?
and also what would i need to do if i want this:
Enter year and month: 1980 2
nineteen hundred eighty<--------------- how do i do this for years from 1792 - 9999???
has three hundred and sixty six (366) days
February has 29 days

So, lets solve one problem at a time. Here is an example of a if/else/if/else chain:

Code:

---

if (x == 1)
    printf("x is 1");
else if (x == 2)
    printf("x is 2");
else
    printf("x is not 1 or 2");

---

I'm not going to write the code for you, because you are not learning anything from that.

As to making "number into text", it's not terribly complicated, but you post something that starts us off, and we'll help you along - that's the "Homework policy" here on Cprogramming.com...

--
Mats

THIS is what i got so far... is it right?

#include <stdio.h>

int main()

{
int year, month;

printf("Enter year and month:");
scanf("&#37;d", &year);
if (((year%4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&& (year%4==0)))
{
printf("has three hundred and sixty six (366) days\n");}
else{
printf("has three hundred and sixty five (365) days\n");}

scanf("%d", &month);
if (month==1);
printf("Janurary has 31 days");}
else if (month==2)
printf("Feburary has 28 days");}
else (month==3)
printf("March has 30 days");}

return 0;
}

Does it compile?

--
Mats

Do not put ending brackets on the same line as other code.
Do use code tags when posting code.

Code:

---

#include <stdio.h>

int main()
{
        int year, month;

        printf("Enter year and month:");
        scanf("&#37;d", &year);
        if (((year%4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&& (year%4==0)))
        {
                printf("has three hundred and sixty six (366) days\n");
        }
        else
        {
                printf("has three hundred and sixty five (365) days\n");
        }

        scanf("%d", &month);
        if (month==1);
        printf("Janurary has 31 days");
}
else if (month==2)
printf("Feburary has 28 days");
}
else (month==3)
printf("March has 30 days");
}

return 0;
}

---

Quote:

---

Originally Posted by Elysia

Do not put ending brackets on the same line as other code.
Do use code tags when posting code.

---

Actually that's not really needed. Out of all the IDE's I've used Visual Studio's auto-formatting just forces you to do that, but putting them at the end of an if statement or onto the next line is just a matter of style.

Quote:

---

Originally Posted by bobbie18

THIS is what i got so far... is it right?

Code:

---

#include <stdio.h>

int main()

 {
    int year, month;

    printf("Enter year and month:");
    scanf("%d", &year);
    if (((year%4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&& (year%4==0)))
{
        printf("has three hundred and sixty six (366) days\n");}
    else{
        printf("has three hundred and sixty five (365) days\n");}

    scanf("%d", &month);
    if (month==1);
        printf("Janurary has 31 days");}
    else if (month==2)
    printf("Feburary has 28 days");}
    else (month==3)
    printf("March has 30 days");}

    return 0;
}

---

---

I was under the impression that March always had 31 days.

Slight improvement on using enumerations as this makes the code slightly self defining.

Code:

---


#include <stdio.h>

int main()
{
        int year, month;
        enum months{January = 1, February, March};

        printf("Enter year and month:");
        scanf("%d %d", &year, &month);
        if (((year%4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&& (year%4==0)))
        {
                printf("has three hundred and sixty six (366) days\n");
        }
        else
        {
                printf("has three hundred and sixty five (365) days\n");
        }

  if (month == January)
    printf("Janurary has 31 days");
  else if (month == February)
    printf("February has 28 days");
  else if (month == March)
    printf("March has 30 days");

return 0;
}

---

Quote:

---

Originally Posted by jake123

Actually that's not really needed. Out of all the IDE's I've used Visual Studio's auto-formatting just forces you to do that...

---

Which is a good thing. But other IDEs are not so nice.

Quote:

---

Originally Posted by jake123

Actually that's not really needed. Out of all the IDE's I've used Visual Studio's auto-formatting just forces you to do that, but putting them at the end of an if statement or onto the next line is just a matter of style.

---

You can write code any way you like from the COMPILERS standpoint. Elysia's comments are intended to make the code more readable, and for a novice that has little experience, the "it's just a matter of style" is probably not helpful. Giving strict, simple, guidance helps a whole lot more. [Shame I don't always follow my own advice].

Quote:

---

I was under the impression that March always had 31 days.

---

So was I.

--
Mats

Lol, yes March has 31 days. I should have used the knuckle method to find that out.

you guys are all blooldy legends first of all and thx for the enum method slingerland3g
but can someone explain to me how the enumeriatsaion method works
and can someone give me some tips on how to convert years into words>??? cause i have absolute no idea

Enumeration is like defining an integer type, but only allowing the values 0, 1, 2. If the user tries to assign anything other than a 0, 1, 2, you get an error.

Code:

---

typedef enum
{
        ENUMERATION_1,
        ENUMERATION_2,
        ENUMERATION_3
} enumeration;

enumeration a;
enumeration b;

---

It's useful for assigning descriptive names to certain values that a variable can take.

so since i have to create a calender from january to decemeber
enumeration wont work eh cause it only works for intergers for 0, 1, 2 but not 3 onwards...
so what other methods can i use if i did to include integers from 0 to 12

Quote:

---

cause it only works for intergers for 0, 1, 2 but not 3 onwards

---

Why do you say so? enum could include any integer you want

Quote:

---

integers from 0 to 12

---

Why do you need 13 months?

for marking an error code

hi guys im just wondering how to make an else if comment this is what i got so far....
if (month == January)
printf("Janurary has 31 days\n");
else if ((month == February)&& (((year&#37;4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&& (year%4==0))))
{ printf("February has 29 days\n");
}
else
{ printf ("February has 28 days\n");
}


However it does not work

use switch

Code:

---

enum months {Jan,Feb,Mar,...} month;
switch(month)
{
  case Jan: ...
      break;
  case Feb:
...
}

---

u mind explaining briefly about how switch works plz?

http://www.cprogramming.com/tutorial/lesson5.html

Quote:

---

Originally Posted by bobbie18

hi guys im just wondering how to make an else if comment this is what i got so far....
if (month == January)
printf("Janurary has 31 days\n");
else if ((month == February)&& (((year%4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&& (year%4==0))))
{ printf("February has 29 days\n");
}
else
{ printf ("February has 28 days\n");
}
However it does not work

---

Post code within code tags! and indent properly.

Code:

---

if (month == January) {
    printf("Janurary has 31 days\n");
} else if ((month == February) &&
          ( ((year%4 == 0 ) && (year%100 != 0)) ||
            ((year%400 == 0) && (year%100 == 0) && (year%4 == 0))
          )
          ) {
    printf("February has 29 days\n");
} else {
    printf ("February has 28 days\n");
}

---

Also use braces with all if/else statements or at least don't mix an if statement without braces with an else statement with braces!
I've also re-aligned your if statement a bit, so you can see whether you're doing what you're intending to do.

QuantumPete

hi i sort of fixed it but don't know what i did wrong when i want the printf statement to be Feb has 29 days when it's a leap year... Code as follow: THX A LOT GUYS

switch(month)
{
case January:
printf("January has 31 days\n");
break;
case February:
printf("February has 28 days\n");
break;
case (((year&#37;4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&&
(year%4==0))) February:
printf("February has 29 days\n");
}

You need to put the condition to examine if it's a leap-year inside the case for February:

Code:

---

if (leap_year) ... 29
else ... 28

---

The case-label, e.g. Februry, must be constant, it can not be "an equation" - it MUST be something that the compiler can figure out at compile time into a simple number.

--
Mats

This might work by repetitive instruction:
use code tags.

http://cboard.cprogramming.com/showthread.php?t=25765

QuantumPete

Code:

---

#include <stdio.h>

int main()
{
        int year, month;
        enum months{January = 1, February, March, April, May, June, July,
        August, September, October, November, December};

        printf("Enter year and month:");
        scanf("%d %d", &year, &month);
        if (((year%4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&&
        (year%4==0)))
        {
                printf("has three hundred and sixty six (366) days\n");
        }
        else
        {
                printf("has three hundred and sixty five (365) days\n");
        }

switch(month)
{
  case January:
    printf("January has 31 days\n");
break;
  case February:
    printf("February has 28 days\n");
break;
  case (((year%4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&&
        (year%4==0))) February:
    printf("February has 29 days\n");
}

return 0;
}

---

OK. Your case label is wrong. Case only accepts either a number of a constant. So you can say case February, because February = 2, which is how you defined your enum.
Now, think about this: when you're inside your February case statement. You're printing that Feb has 28 days at the moment. How about putting your leap-year check within the case statement, and then print out either 28 or 29 depending?

QuantumPete

thx for the reply
but how do i place the leap-year check within the case statement.
do i have to define leap yr from the beginning?

Inside your February case, use an if to determine if it's a leap year or not and print appropriate information. Shouldn't be difficult.

thx elysia for u help again
this is for i got so far... where should i put the if statement for leap yr? cause the below codes dont work when i compile it

Code:

---


switch(month)
{
  case January:
    printf("January has 31 days\n");
break;

  if (((year%4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&&
        (year%4==0)))&& case  February:

    printf("February has 29 days\n");

  else

    printf("February has 28 days\n");

}
return 0;
}

---

Well, of course it won't.
If you think logically... make an internal flowchart.
Where does the leap year code need to be put?

sorry i have only learnt c for a couple of days
whatz an internal flowchart?

It means that you design or write down how want you want to perform flows.
So, for example, in your example, a simple flowchart might be:

Read input from user
...
Close program

By writing down each step, you can easily know where to place code.

C programs are a sequence of instructions to do things. A flowchart is a diagramatic way to describe how a program works.

But you don't need to have a flowchart - you can use many other ways to describe the code. What is important is that you understand that the compiler has very strict rules about what goes where, and how the code can and can't be organized, and the sequence of your code is important. If you want to bake a cake, you have to do things in order: turn on the oven, mix the ingredients, pour the mix in a cakeform, then put it in the oven, wait a specified amount of time, then take the cake out and turn off the oven. If you put the raw ingredients in the oven, then mix them afterwards, it won't be a cake - it may still turn out to be sort of edible, but chances are that it won't be very nice.

Same applies to programs - your program is a RECIPE for doing some particular calculations or work.

--
Mats

ah k thx i sort of think logically for a while but my method still don't work (sorry im dumb)

Code:

---

switch(month)
{
  case January:
    printf("January has 31 days\n");
break;

    case  February:
      printf("February has 28 days\n");
 
    else if (((year%4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&&
        (year%4==0)))

    printf("February has 28 days\n");
 
}
return 0;
}

---

You have "else if" without if.
How can you "do something else" without "something" first?

any tips at all about where i need to put or define the leap year code in the program?
because im really lost

You're on the right track, but you're still making a hash of it.

You need to start with an if, not an else. And you probably want that FIRST, then print afterwards - like cake-baking, you're currently putting the ingredients in first (printing 28 days), then mixing (checking if it's a leapyear).

--
Mats

Code:

---

enum months
{
        January = 1,
        February,
        March,
        April,
        May,
        June,
        July,
        August,
        September,
        October,
        November,
        December
};

int main()
{
        int year;
        enum months month;
        printf("Enter year and month:");
        scanf("%d %d", &year, &month);
       
        switch(month)
        {
        case January:
                printf("January has 31 days\n");
                break;
        case February:
                if (((year%4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&&
                        (year%4==0)))
                {
                        printf("February has 29 days\n");
                }
                else
                {
                        printf("February has 28 days\n");
                }
                break;
        case March:
                printf("March has 31 days\n");
                break;
        }
       
        return 0;
}

---

thx mats
but i thought i need to get the program to check whether the input before for the year is leap or not
then decide whether feb is 28 or 29 days

Code:

---

//Create by z3252440
#include <stdio.h>

int main()
{
        int year, month;
        enum months{January = 1, February, March, April, May, June, July,
        August, September, October, November, December};

        printf("Enter year and month:");
        scanf("&#37;d %d", &year, &month);
        if (((year%4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&&
        (year%4==0)))
        {
                printf("has three hundred and sixty six (366) days\n");
        }
        else
        {
                printf("has three hundred and sixty five (365) days\n");
        }

switch(month)
{
  case January:
    printf("January has 31 days\n");
break;

    if (((year%4==0 )&& (year%100!=0)) || ((year%400==0) && (year%100==0)&&
    (year%4==0)))

    case February:

    printf("February has 28 days\n");

    else

    printf("February has 29 days\n");
break;
}

return 0;
}

---

As can seen above shouldnt the program check the year then decide which statement to print?
cause this is where i am getting really confuse

see my sample

Read your code. Each and every line and understand exactly what it does.
Then write a flowchart with that, knowing what it does.
Then insert your check where appropriate in the flowchart.
Then translate it to code.

It will do you wonders.

varts what can i say you are a legend
so the real devils are { and }
thx for the help mats and elysia
since im on it
im just wondering how do i convert year outputs into words?
does it involve with enumerations cause if it does dont i have to create some sort of dictionary?
any pointers plz?

You will need some sort of translation of number to string - if you want to call that a dictionary, then that's probably closer than an "enumeration".

--
Mats

Quote:

---

Originally Posted by vart

see my sample

---

that's not a sample, that's the answer :P

QuantumPete

translation of number to string....
right....
are there websites or tutorials that can tell me about that cause i have no idea how to start at all
thx mats

P.s: does it invole #include <stdlib.h>??? there's only one setence from this website (http://www.cs.cf.ac.uk/Dave/C/node16...00000000000000)
which does not really help

Don't worry about including headers.
Worry about what functions to use, and when you use those functions, include appropriate header.

haha i just need a header in case my teacher saw this thread and said i plagiaries

You definitely don't need stdlib.h for this particular task:

Code:

---

#include <stdio.h>

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

  int n;

  printf("Enter a number:");
  fflush(stdout);
  scanf("%d", &n);
  printf("Last digit of number is %s", numbers[n % 10]);
  return 0;
}

---

This solves PART of your problem - now you have to find the rest of the solution.

--
Mats

Including a header without using is bad. Especially if you just want someone to "see" that header.

thx for the sample mat
can someone explain what is
fflush(stdout);?
and
char *numbers[] ?

fflush(stdout) -> flushes output buffer, makes sure the user sees the "Enter a number:" text.
char* numbers[] (as I like to type it) clearly shows that it's an array of pointers (type) and [] means that the dimensions are undefined (the compiler will determine this depending on what you initialize it with).

But this should really be const char*...

Quote:

---

Originally Posted by QuantumPete

that's not a sample, that's the answer :P

QuantumPete

---

Now that he's got the answer, I'd like to post my version

Code:

---

void processInput(int year, int month)
{
    int leap = ( (year%4 == 0 && year%100 != 0) || year%400 == 0 )? 1:0;
                // one of the ways to check if a given year is leap
               
    int monthDays[] = {31,28+leap,31,30,31,30,31,31,30,31,30,31};
                // adjust num of days in February accordingly
               
    const char *monthStr[12] = {"Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"};
                // string constants for each month
               
    char yearStr[40];
                // string variable for year in words

    getYearInWords(yearStr, year);

    printf("year %d (%s) has %d days\n", year, yearStr, (365+leap));
    printf("month %d (%s) has %d days\n", month, monthStr[month-1], monthDays[month-1]);
}

---

I've got the function

Code:

---

getYearInWords(yearStr, year);

---

ready but didn't post it coz i don't want to stop him learning.
Anyway, this function just uses logic as explained earlier by matsp

cheers
maverix

But if you want something fancy, you can do this ;)

Code:

---

int leap = !!( (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0) );

---

Quote:

---

Originally Posted by Elysia

But if you want something fancy, you can do this ;)

Code:

---

int leap = !!( (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0) );

---

---

You don't need !! for that, as it' produces a "bool" in the first place.

--
Mats

You may be correct...
I didn't really think of it.

hi guys im back
i just realised for the coverting years into words bit like eg:
2008 ----> twenty hundred and eight
How do i do it if i am only allowed using the if method?
so far i got
year divide by 1000 and then using the remainder but i sort of lost myself on that... any more suggestions?

Something like
2008 / 1000
2008 / 100 & 0x9
2008 / 10 0x9
2008 & 0x9
Might work for getting each decimal

Quote:

---

Originally Posted by Elysia

Something like
2008 / 1000
2008 / 100 & 0x9
2008 / 10 0x9
2008 & 0x9
Might work for getting each decimal

---

And'ing with 0x9 won't work since that will only peel off 1 and 8. Use remainder from the previous division instead.

Bah, it was worth a try.

If you want "twenty hundred and eight", then you probably want to divide by 100, rather than by 1000 - if you want "two thousand and eight", you should divide by 1000.

--
Mats

can someone please demonstrate a code of how to do it for years 4321 for eg. and also i have ABSOLUTE no idea on how to do years such 1700 to 1999. can someone show me a code. Im not getting u guys to do my work but im really lost on how to do this one

my teacher forced us to make a program from the stuff we learn such as if statements. so i think i have to create at least 49 if statements but can someone plz show me an example plesase. BEgging you i got so much other work i got to catch up with and this program is burning my time away

An example of what, exactly?

--
Mats

a sample of the code....
i got this so far

Code:

---

    if (year/1000)=4
        printf("forty")
        if ((year%1000)/10)=3
        printf("three hundrend")
        if ((year%1000)/100)=2
        printf("twenty")
        if ((((year%1000)/100)

---

but i dont know how to finish it or do years such as 1822 for eg

Sample:

Code:

---

        int year = 2008;
        char buf[5];
        char buf2[100] = {0};
        sprintf_s(buf, sizeof(buf), "%i", year);
        char thousand = buf[0];
        char hundred = buf[1];
        char tenth = buf[2];
        char oneth = buf[3];
        if (thousand == '2') strcat_s(buf2, sizeof(buf2), "Two thousand");
        strcat_s(buf2, sizeof(buf2), " and ");
        //if (buf[1]
        if (oneth == '8') strcat_s(buf2, sizeof(buf2), "eight");
        puts(buf2);

---

If the compiler doesn't accept strcat_s and sprintf_s, then just use strcat and sprintf.

elysia thx for ur reply...but i think this too compilcated. we are only allowed to use if and else if statements... is there any chance u can make this simpler?

It's as simple as it will be unless you involve a lot of math.
It just converts it to a string and takes the first number, the second, third and fourth.
So the 2, 0, 0, and 8, and translates them into a word using ifs.

First of wall, you need to decide if you want twenty-hundred or two-thousand.

If you can't use anything beyond if-statements [which is pretty limiting for this sort of excercise], then it would help to be able to use a function - but you probably aren't allowed that either.

So, divide the number into portions FIRST, like "hundreds = year / 100", "thousands = year / 1000", "tens = (number / 10) % 100", "ones = number % 10". Then you need a big-if-tree where you split decide which number to print based on the thousands, hundreds, etc, etc.

If you can use functions, you can simplify it by making a function that prints "zero" to "ninetynine", and then call that to print thousands, hundreds, and "the rest".

And you can of course use arrays if that's allowed, like in my example I posted yesterday.

--
Mats

i got the following but i do not know why it doesn't work

Code:

---

{
        if ((year/1000)=4)
        {
        printf("forty");
        }
        if (((year%1000)/10)=3)
        {
        printf("three hundrend");
        }
        if (((year%100)%100)=2)
        {
        printf("twenty");
        }
        if ((year%10)=1)
        {
        printf("one\n");
        }
}

---

Quote:

---

Originally Posted by bobbie18

i got the following but i do not know why it doesn't work

Code:

---

{
        if ((year/1000)==4)
        {
        printf("forty");
        }
        if (((year%1000)/100)==3)
        {
        printf("three hundrend");
        }
        if (((year%100)/100)==2)
        {
        printf("twenty");
        }
        if ((year%10)==1)
        {
        printf("one\n");
        }
}

---

---

And of course, that only works for the year 4321.

i c i c thx for the reply
can you give me some ideas of how to write the year 1782 for eg?
cause wont i have to create a eleven to nineteen dictionary for the program?
any samples?

and i dont know why the output is
fortythree hundrendone
i cant get the twenty out for some reasons???

Quote:

---

Originally Posted by bobbie18

i c i c thx for the reply
can you give me some ideas of how to write the year 1782 for eg?
cause wont i have to create a eleven to nineteen dictionary for the program?
any samples?

---

You'll need every word you'll want to print. Seventeen, sixty, eight, forty, etc. The idea hasn't changed since you got it right two posts ago. Divide.

but let say 1782
how does the divison work on that?
1782/1000 and then ???
lol im so confused
PLEASE do you mind giving me an example
PLEASE literally begging you

Quote:

---

Originally Posted by bobbie18

but let say 1782
how does the divison work on that?
1782/1000 and then ???
lol im so confused
PLEASE do you mind giving me an example
PLEASE literally begging you

---

Missed one: that should be a divide by 10 in the twenty, not a divide by 100.

I don't know what else to say except you've already posted the example. But just because:

Code:

---

int number = 4321;
int thousands = number/1000;
int hundreds = (number%1000)/100;
int tens = (number%100)/10;
int ones = number%10;

if (thousands == 1)
    printf("One");
else if (thousands == 2)
    printf("Two");
else if (thousands == 3)
    printf("Three");
else if (thousands == 4)
    printf("Four");
else if (thousands == 5)
    printf("Five");
else if (thousands == 6)
    printf("Six");
else if (thousands == 7)
    printf("Seven");
else if (thousands == 8)
    printf("Eight");
else if (thousands == 9)
    printf("Nine");
if (thousands)
    printf(" thousand ");

---

et cetera.