why its not working?

[Dulus]

hi
having an irritating problem. i've tried everything i know, still, cant figue why its not working..

the program should get from you a "n" value, and calculate the deniz function as clearly be seen...

Need a little help please..

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
long deniz ( double x);

int main(void)
{
  int i;
  int n;
  long total = 0;
  printf (" n value ?? \n");
  
  scanf ( "%d", &n );
  
   for ( i = 1;i <= n; ++i )
   {  total = total + deniz(i);
   
}
printf( "output %f", total);
  system("PAUSE");	
  return 0;
}
long deniz (double x)
{

return log(x)-x ;
}

[Prelude]

>for ( i = 1; 1 <= i <= n; ++i )
That's not meaningful. I suspect you meant:

for (i = 1; i <= n; ++i)

[Dulus]

ah. figures...

yep, thanks. i've corrected it. i dont know why i wrote that...

but still :S

when i run the program, says

"output 0.000000"

[Prelude]

You're using %f to print a long integer. Try %ld instead, or change total to either float or double.

[jimblumberg]

Check you format specifier for the following printf:

printf( "output %f", total);

Is total a double?

Jim

[Dulus]

right...
very very thank you again.
i got to study a little more i guess...

gentleman,
i still got a problem.
the program still doesent give expected output.

for example, for n=3 output has to be -4,208240...

but it gives me -3...

its sad...

[jimblumberg]

You are mixing longs with doubles. Long like int will truncate the result to an int value.

You should change to using double and not long.

Jim

[CommonTater]

One possibility... Your i variable is an integer but the deniz function is asking for a double.

You should use a typecast when calling the function...

total = total + deniz((double) i);

Also... please don't go back and edit your code. Leave the old one there and post a new copy so that others with the same problem can follow how you solved it.