for example if i input 2 floating numbers
number 1 : 5.7
number 2: 7.3
and i want to comapre them BUT i only want to comapre fractional part
so number 1 would be bigger than number 2 because fraction of number 1 is bigger. (7 > 3)
so My question is...how do i compare fractions only? Would formating the number work? If after inputing a number 1 i formated it to .7 ...would it work like that?
I think modf is standard in math.h, this should split the float, you'd then be able to do your comparison.
Demonographic rhinology is not the only possible outcome, but why take the chance
You can use the modf() function to give you the fractional part.
Code:#include <stdio.h> #include <math.h> int main(void) { double a = 5.7; double b = 7.3; double ipart; if (modf(a,&ipart) > modf(b,&ipart)) printf("fractional part of a is greater than b.\n"); return 0; }
Hi Mak
The answer to your question is NO, formating had nothing to
do with the internal representation of the numbers, You see, formating only afects how WE humans see the numbers, internaly they are someting else,
If you want to compare the fractions, you had to isolate them,
all you had to do is substract the integer part, and compare them
hope this was useful
Saludos desde Mexico
Chuky
Converting from a floating point value to an integral value the fractional part is discarded. So if you subtract off the integral portion, you would be left with the fractional part. So, here is one way.[edit]Damn I was poky on this one!Code:#include <stdio.h> int main(void) { double n1 = 5.7; double n2 = 7.3; int i1 = (int)n1; int i2 = (int)n2; double f1 = n1 - i1; double f2 = n2 - i2; if(f1 > f2) { printf("%g > %g\n", f1, f2); } return 0; } /* my output 0.7 > 0.3 */
Last edited by Dave_Sinkula; 10-14-2003 at 07:31 PM.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
why is the "ipart" there?Originally posted by swoopy
You can use the modf() function to give you the fractional part.
Code:#include <stdio.h> #include <math.h> int main(void) { double a = 5.7; double b = 7.3; double ipart; if (modf(a,&ipart) > modf(b,&ipart)) printf("fractional part of a is greater than b.\n"); return 0; }
and if i try using modf...i get an error:
warning: passing arg 2 of `modf' from incompatible pointer type
and does it matter if i use float insted of double?
okay...here is my first try:
Code:#include <stdio.h> int main(void){ float num; float num2; int i1, i2; printf("Please enter a float number : "); scanf("%f",num); /*GETS FIRST FLOAT NUMBER */ while (num !=0) /*MAKES SURE ITS NOT 0 */ { printf("Please enter a float number : "); scanf("%f",num2); /*GETS SECOND FLOAT NUMBER */ /*turns it into fractions only*/ i1 = (int)num; i2 = (int)num2; num = num - i1; num2 = num2 - i2; /*compares and prints out the bigger one*/ if(num>num2) { printf("The number is %f.\n",num); num=num2; } if(num2>num) { printf("The number is %f.\n",num2); num2=num; } } printf("DONE"); }
now for some reason that i cant understand i get "Access violation" error right after i enter first float number ....
okay...it works now.
I just need to work on the logical part now.....it does not work as it should.
