Casting float to int in C++

Question

int uniquePaths(int m, int n) {
    int num = m+n-2;
    int den=1;
    double ans = 1;
    while(den<=m-1) {
        ans = ans*(num--)/(den++);
    }
    cout<<ans;
    return (int)ans; 
}

The expected answer for m=53, n=4 as input to the above piece of code is 26235 but the code returns 26234. However, the stdout shows 26235.

Could you please help me understand this behavior?

What do you think casting a `float` (actually, a `double`) to an `int` is supposed to do? — Scott Hunter, Jan 13 '23 at 19:25
Find out the *exact* value of `ans` (hint: print `ans-26235`). — Scott Hunter, Jan 13 '23 at 19:32
Does this answer your question? [Is floating point math broken?](https://stackoverflow.com/questions/588004/is-floating-point-math-broken) — Kevin, Jan 13 '23 at 20:45

Eric Postpischil · Accepted Answer · 2023-01-13T22:32:54.023

Due to floating-point rounding, your code computes ans to be 26,234.999999999985448084771633148193359375. When it is printed with cout<<ans, the default formatting does not show the full value and rounds it to “26235”. However, when the actual value is converted to int, the result is 26,234.

After setting num to m+n-2, your code is computing num! / ((m-1)!(num-m+1)!), which of course equals num! / ((num-m+1)!(m-1)!). Thus, you can use either m-1 or num-m+1 as the limit. So you can change the while line to these two lines:

    int limit = m-1 < num-m+1 ? m-1 : num-m+1;
    while(den<=limit) {

and then your code will run to the lower limit, which will avoid dividing ans by factors that are not yet in it. All results will be exact integer results, with no rounding errors, unless you try to calculate a result that exceeds the range of your double format where it is able to represent all integers (up to 2⁵³ in the ubiquitous IEEE-754 binary64 format used for double).

Casting float to int in C++

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