double integer math problems surpass my programming skills

ok heres the code:

Code:

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

int main (int argc, char *argv[])
{
	double d; int j;

	d = (13 % 4);
	printf("rint : d = %f\r\n", d);
	j = rint(d);
	printf("rint : d = %f, j = %d\r\n", d, j);
	d = (13 % 4);
	printf("round: d = %f\r\n", d);
	j = round(d);
	printf("round: d = %f, j = %d\r\n", d, j);
}

heres the output:

Code:

rint : d = 1.000000
rint : d = 1.000000, j = 1
round: d = 1.000000
round: d = 1.000000, j = 1

How come d doesn't equal 3.25 and then how come d doesn't equal 3 or four depending on the round function?

Originally Posted by errigour

How come d doesn't equal 3.25

Why should d be 3.25? You assign (13 % 4) to it, and 13 % 4 == 1.

Originally Posted by errigour

d doesn't equal 3 or four depending on the round function?

You're not assigning the result of round(d) to d anyway.

I thought 13/4 equals 3.25?

and how come when i do d = (13 / 4); d only equals 3 even before rounding d.

13 / 4 is done in integer maths, THEN the result is converted to a double and then assigned to d.

13.0 / 4.0 is done using floating point maths, and then assigned to d.

Originally Posted by errigour

I thought 13/4 equals 3.25?

13/4 and 13%4 are very different expressions. Look at which one you actually used.

Thanks Salem. So what the hell does 13 % 4 do?

I have never understood the % sign does maybe you could explain in to me in a way that I could understand, or not at all.

Originally Posted by errigour

So what the hell does 13 % 4 do?

It results in the remainder of 13 divided by 4.

so 13.0 % 4 = .25? if yes thanx if not i'm lost. Thanx!

Originally Posted by errigour

13.0 % 4 = .25?

No, because % is not defined for floating point numbers. Rather, #include <math.h> and use fmod, e.g.,

Code:

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

int main(void)
{
    printf("%f\n", fmod(13.0, 4));
    return 0;
}

Originally Posted by errigour

so 13.0 % 4 = .25? if yes thanx if not i'm lost. Thanx!

Code:

13 % 4 == 1

Modulus (%) gives the remainder of division:

------------
  ____
4|13

------------
   3
  ____
4|13

------------
   3
  ____
4|13
  12
  ----
   1  <-- remainder
------------

hey is there any way to get the value of a division of integers into a double etc...
int i, j;
double b;
i = 14;
j = 4;
b = i / 4;

Cast one of the parts to a double first:

Code:

b = (double) i / 4;

What happen's to integers when the sum is never ending?