Compare integer and float

Question

I have used this code to generate root, if output is not integer it should return -1.

#include <stdio.h>
#include <math.h>
int main(){
    long int n=0;
    long int num,p,reti;
    double ret;
    scanf("%ld",&n);
    while(n!=0){
        scanf("%ld %ld",&p,&num);
        ret=pow(num,(double)1/p);
        reti=(int)ret;
        printf("%lf %lf\n",ret,(double)reti);
        if(ret!=(double)reti){
            reti=-1;
        }
        printf("%ld\n",reti);
        n=n-1;

    }
    return 0;

}

but if I give input as

1
5  3125

it should give 5, but it is giving -1.

Please reformat your post so we can read the code in a code block. Ie add spaces in front to each code line! — Mike Wodarczyk, Feb 12 '17 at 05:22
keep variable names understandable !what is n,p,num...why not keep it like number , power — minigeek, Feb 12 '17 at 05:28
what does the earlier printf line show? I would suspect that the pow function returns something close to 5 but not exactly 5.0000000000. (eg 5.00000000001) Then when you (int) it and get 5 and convert this back to a double you see that 5.0000000000 != 5.00000000001. — Mike Wodarczyk, Feb 12 '17 at 05:34
1/5 is not exactly representable in binary floating-point, so even if you assume that `pow()` will produce an answer within one half ULP of the exact one *for the computation you are performing*, that answer may not be exactly 5, because the power is not exactly 0.2. — John Bollinger, Feb 12 '17 at 05:35

score 1 · Answer 1 · edited May 23 '17 at 12:01

ret is not exactly 5.00000, you're just not printing enough digits after the decimal point to see the difference. Change your format to %.20lf and it will show 5.00000000000000088818, which is not equal to (double)5.

This is because pow() works with floating point numbers, and there are some inaccuracies introduced. (double)1/5 can't be represented exactly in binary floating point. See Is floating point math broken?

score 0 · Answer 2 · answered Feb 12 '17 at 06:01

The reason why is this block of code:

if(ret!=(double)reti){
    reti=-1;
}

reti is some integer value - converting it to double will convert it to a double representation, but odds are good that it will not be exactly equal to ret. This is a common issue with floating point numbers - generally speaking, you should never compare pure equality. Instead, you should compare the difference between the two numbers and confirm that it's within some acceptable error range (e.g. 0.00001).

Compare integer and float

2 Answers2