I have a call to
long long a = sqrt(n/2);
Both a
and n
are long long
's but it won't let me compile because it says my use of sqrt()
is an ambiguous call. I don't see how it's possibly ambiguous here at all. How do I resolve this? I have the same problem with floor()
.
My includes
#include "stdafx.h"
#include <iostream>
#include <cmath>
using namespace std;
There are several overloads of sqrt()
and floor()
, there's no "best match" for a call to sqrt(long long)
according to the overload resolution rules. Just cast the argument to the appropriate type -- i.e.,
long long a = sqrt(static_cast<double>(n/2));
//use
sqrt(static_cast<double>(n/2));
//instead of
sqrt(n/2);
The sqrt
functions expects a float
, a double
or a long double
:
long long a = sqrt(n * 0.5);
You may lose some precision converting a long long
to a double
, but the value will be very close.
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