Converting double to int is giving unexpected results

Question

I was trying to find cube root of a perfect cube using pow function but the answers were unexpected (3375 = 15^3)

#include <bits/stdc++.h>

using namespace std;

int main()
{
    double cb = pow(3375, 1.0/3.0);
    printf("cb(in double) = %lf\n", cb);
    printf("cb(in int) = %d\n",(int)cb);
    return 0;
}

the output it shows is :

cb(in double) = 15.000000
cb(in int) = 14

After discussing it with people it turned out that the code ,

printf("%0.0lf", 14.0/3);

gives output

I am not getting why it's happening, and if it's due to precision in storing double and rounding off, then it should round it down to a smaller value rather than rounding to a greater value.

If what you saying i scorrect then why 14.0/3 is giving output as 5 — prashant shishodia, Jun 28 '18 at 17:17
Perhaps in your machine `cb` results in something a bit over 15, say 15.000000003 which rounds to int 15. Or it's 14.9999999966? — Ripi2, Jun 28 '18 at 17:18
[Why should I not #include ?](https://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h) — Jesper Juhl, Jun 28 '18 at 17:18
14.0/3 yields 4.666666. When you round 4.666666 up you get 5. — Eljay, Jun 28 '18 at 17:20
You should use your debugger to see the actual value instead of using `printf`. — PaulMcKenzie, Jun 28 '18 at 17:21
@PaulMcKenzie it's giving the same answer with cout and setprecision(0) — prashant shishodia, Jun 28 '18 at 17:23
4.6667 rounded to 0 decimal places is 5 – it isn't obvious why you're puzzled. What else would you expect? Assignment is different from printing. And `printf("%d\n", (int)(14.0/3));` would yield 4. — Jonathan Leffler, Jun 28 '18 at 17:27
@JesperJuhl it's giving the same output on using cstdio cmath and iostream as headers instead of bits/stdc++.h — prashant shishodia, Jun 28 '18 at 17:27
@prashantshishodiap The point is that you are relying on output functions instead of viewing the actual value. That is not a good way to determine the actual value. Use the debugger. — PaulMcKenzie, Jun 28 '18 at 17:27

ash108 · Accepted Answer · 2018-06-28T17:33:25.730

TL;DR: (int)double always rounds towards zero. printf("%0.0lf", double) actually rounds to closest integer.

When you convert the double to int, then the result is rounded towards zero. E.g. (int)14.9999 is 14, not 15. Therefore your first example with pow() must have given you the result slightly below 15, therefore direct cast to int gives 14.

Now, when you use printf(), then other rules are used. Printf, when asked to print a double using smallest number of digits possible (%0.0lf) rounds it to the closest integer, therefore printf("%0.0lf", 14.666) prints 15.

Converting double to int is giving unexpected results

1 Answers1