Thread: My very first own C program (I need a hand!)

thanks all!!
I think some algebra needed to be rethought
Here's the right code:

Code:

#include <stdio.h>
int main()
{
       float e,x,y,z;
       printf("Enter value for x :");
       scanf("%f",&x);
       y = (60/x);
       e = (y/2);
       z = e*1000;
       printf("Echo = %f\n",z);
       getch();
       return 0;
}

Well done.

One last thing, you should only declare main in one of two ways:
int main( void ) /*When it takes no arguments.*/
int main( int argc, char **argv ) /*When it takes arguments from the command line*/

In your case you want the first one, you will probably learn about the second one later on.

It will still work if you use 'int main()' but it's good practice to get things right from the start, so you don't have to relearn them later.

Originally Posted by DeadPlanet

Well done.

One last thing, you should only declare main in one of two ways:
int main( void ) /*When it takes no arguments.*/
int main( int argc, char **argv ) /*When it takes arguments from the command line*/

In your case you want the first one, you will probably learn about the second one later on.

It will still work if you use 'int main()' but it's good practice to get things right from the start, so you don't have to relearn them later.

Then I have a question "what is an argument exactly?" does this mean that a value is returned to a different function?

Originally Posted by Exutus

thanks all!!
I think some algebra needed to be rethought
Here's the right code:

Code:

#include <stdio.h>
int main()
{
       float e,x,y,z;
       printf("Enter value for x :");
       scanf("%f",&x);
       y = (60/x);
       e = (y/2);
       z = e*1000;
       printf("Echo = %f\n",z);
       getch();
       return 0;
}

Single letter variables are an excellent way to confuse people. It's bad news that you are apparently done, and I still have no idea, at a glance, what this calculates. Your poor user must be confused as well. What the heck is an X? How could he possibly avoid typing errors with that one?

Originally Posted by JOZZY& Wakko

Then I have a question "what is an argument exactly?" does this mean that a value is returned to a different function?

In common parlance the terms argument and parameter are interchangeable, but the word argument defines the expression you passed in (e.g. foo, foo+4, 42, etc.) and parameter defines the variable you use in the function (e.g. dest, where, coolWindow, etc.).

Originally Posted by whiteflags

Single letter variables are an excellent way to confuse people. It's bad news that you are apparently done, and I still have no idea, at a glance, what this calculates. Your poor user must be confused as well. What the heck is an X? How could he possibly avoid typing errors with that one?

I agree.

im also confused as to why you have variable names x,y,z and e. to me that means the e is special, significant in some way.

also, why do you have 3 lines of arithmetic? as opposed to something like this?

Code:

#include <stdio.h>
int main()
{
       float x;
       printf("Enter value for x :");
       scanf("%f",&x);
       printf("Echo = %f\n", 30000/x);
       return 0;
}

Originally Posted by Brain_Child

also, why do you have 3 lines of arithmetic? as opposed to something like this?

If you want to quibble about that, wait until you calculate something complicated, then you'll rethink measuring good code by line count. I do not think that and (60.0f / x / 2.0f) * 1000.0f can be equivalent, either.

Of course, I'm also assuming integer math is unwanted.

Originally Posted by whiteflags

If you want to quibble about that, wait until you calculate something complicated, then you'll rethink measuring good code by line count. I do not think that and (60.0f / x / 2.0f) * 1000.0f can be equivalent, either.

I've written some complicated code and I am in no way measuring the quality of this program by line count.

I just dont understand why 3 extra variables have to be created (I also dont understand what x is and what is being calculated)

That is why I dont think it is good code

Originally Posted by Dino

This is wrong, and now I'll give you a chance to correct your statement and redeem yourself.

Hmm, I'm not sure where I went wrong with that?

Hello and thanks for all your replies.
I understand that it does not look user-friendly, and this program is for me, not for other users. But I'll post it simple so you will know what it was all about.

a minute divided by the bpm
(the result divided by 2) * 1000

So I guess it is
((60/BPM)/2)*1000

So I made:
y = 60/bpm
then
e=y/2
then
z=e*1000

Can I do it in a single equation ? It could be z = ((60/BPM)/2)*1000

Can try it now, I'm at work on (lunch time)

Code:

#include <stdio.h>
int main()
{
       float e,BPM,y,z;
       printf("Enter the tempo in Beats Per Minute :");
       scanf("%f",&BPM);
       y = (60/BPM); // The result of a minute (60 seconds) divided by the number of beats per minute
       e = (y/2);
       z = e*1000;
       printf("Set the DELAY to %f miliseconds\n",z);
       getch();
       return 0;
}

Also, there are things I haven't understood yet that you are talking about (arguments ?)