Thread: C Programming Help - 'return' & digits in a number

Originally Posted by jamestgarner

I need some help, so any way you can help me would be greatly appreciated. I've just started learning C language and using user defined functions, but I'm still not quite sure as to what "return ;" does or how to use it. At the bottom of my programs I've been instructed to put "return 0;" and when using user defined functions I need to return a number, but don't know how to return it, where it goes when it is returned, or how to retrieve it in the main. Any information about the return command would be helpful.

Generally like this.

Code:

int foo( /* parameters */ )
{
   int value = 0;
   /* calculate value */
   return value;
}

Originally Posted by jamestgarner

Also, I need to create a user defined function that will keep track of how many digits are in the input that a user puts in. For example: I ask the user to input his salary and collect the input with a scanf function. If the user inputs 30000, I need some way for the program to keep track that 5 digits were inputted. Please help me if you know a way to do this or a command that will help me.

It depends on what your input is. If it is text, you could use strlen. If it is a number, that is a basic exercise using division. I'm lazy, so I might call a standard function to find the log base 10.

I generally tell folks to avoid scanf, and the FAQ shows better methods.

Given the function prototype (same as first line of function, exept with ';'):
int numDigets(int number);

you call it like this:

Code:

int main()
{
     int var=numDigets(30000);
     //var equals 5. 
     return 0;
}

You can also do artithmatic expressions on the return value.
And you can pass a variable instead of 30000 as needed.

Originally Posted by jamestgarner

i can call my own function and use log base 10, but i don't understand log very much and so idk how to set something like that up. how would you do that?

Something like this?

Code:

#include <stdio.h>
#include <math.h>

int digits(int number)
{
   return log10(number) + 1;
}

int main(void)
{
   int i;
   for ( i = 10; i <= 10000; i *= 10 )
   {
      printf("digits(%d) = %d\n", i, digits(i));
   }
   return 0;
}

/*my output
digits(10) = 2
digits(100) = 3
digits(1000) = 4
digits(10000) = 5
*/

No, he's correct. You need to make a special case for zero.

Perhaps even any value less than 1.

Code:

if(number == 0)
{
    return 1;
}
else if(number < 0)
{
    number = -number;
}
//then do the log10