Function e^x returns 'inf' as output when x is larger then 0.6-CodePudding

I tried to implement this code and it works to a certain point (x<0.6). I am just wondering why it ouputs 'inf' although the stop criteria should terminate the program when it reaches the maximum accuracy of double.

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

double fak(int n) {
    int f = 1;
    int i = 0;
    
    do {
        i  ;
        f *= i;
    } while(i<n);
    
    return f;
}

double func_e() {
    double res = 0;
    double res_old = 0;
    double x, k;
    x = 1;
    k = 0;
    
    do {
        res_old = res;
        res  = ((pow(x,k)) / fak(k));
        k  ;
    } while(res != res_old);
    
    return res;
}

int main(void) {
    //printf("power %f", pow(3,3));
    
    printf("%f", func_e());
    //printf("%f", fak(3));
    
    printf("\n");
    return 0;
}

CodePudding user response：

Check the return value of your function fak. It will overflow and at a certain point return 0. The division by 0.0 results in inf.

When I modify function fak as

double fak(int n) {
    int f = 1;
    int i = 0;
    
    do {
        i  ;
        f *= i;
    } while(i<n);
    
    printf("fak(%d) = %d\n", n, f);
    return f;
}

and run it on https://onlinegdb.com/ZxaXfI5xcG, the output is

fak(0) = 1
fak(1) = 1
fak(2) = 2
fak(3) = 6
fak(4) = 24
fak(5) = 120
fak(6) = 720
fak(7) = 5040
fak(8) = 40320
fak(9) = 362880
fak(10) = 3628800
fak(11) = 39916800
fak(12) = 479001600
fak(13) = 1932053504
fak(14) = 1278945280
fak(15) = 2004310016
fak(16) = 2004189184
fak(17) = -288522240
fak(18) = -898433024
fak(19) = 109641728
fak(20) = -2102132736
fak(21) = -1195114496
fak(22) = -522715136
fak(23) = 862453760
fak(24) = -775946240
fak(25) = 2076180480
fak(26) = -1853882368
fak(27) = 1484783616
fak(28) = -1375731712
fak(29) = -1241513984
fak(30) = 1409286144
fak(31) = 738197504
fak(32) = -2147483648
fak(33) = -2147483648
fak(34) = 0
fak(35) = 0
inf

This means your loop ends when both res and res_old have the value inf.

Additional remark:

In func_e you use double k; and pass this to double fak(int n) which converts the value to int. Function fak does the calculation in int and implicitly converts the result to double in the return statement.

I suggest to avoid these conversions. (Or at least think about the possible problems.) The compiler may warn about this if you enable all warnings.