Thread: log10(long long) ?

I have a long long type that i am trying to find the log10 of but I can't seem to get around an error..any solutions?

thanks

(n2 is long long)

Code:

unsigned long long L2 = log10(n2) /  log10(2);

Code:

Error	1	error C2668: 'log10' : ambiguous call to overloaded function	c:\documents and settings\all users\desktop\3216968-src.cpp	14

log10 - C++ Reference

takes double and long double..should I be trying to do a conversion?

(and if it matters, the value i am holding in n2 is slightly under 1million)

thanks

Be warned that long long and unsigned long long are just compiler extensions at this point of time, although they will likely be part of the next version of C++.

Originally Posted by rodrigorules

takes double and long double..should I be trying to do a conversion?

That would indeed resolve the ambiguity:

Code:

unsigned long long L2 = static_cast<unsigned long long>(log10(static_cast<double>(n2)) / log10(2.0));

Originally Posted by EVOEx

If you're using "using namespace std"... then try "::log10" in stead of log10. Hooray for namespace pollution.

This is not reliable: it may well be the case that there are overloads of log10 in the global namespace with parameter types that match a long long equally well.

Originally Posted by EVOEx

DON'T use using namespace std.

In this context, such a using directive should be okay.

If you're working with long long, you may consider using long double for the log input, for more precise calculation.

Since you are looking for a result that is integral (i.e. have to round up or down anyway) it isn't hard to compute the logarithm directly. For example (assuming your compiler supports long long types);

Code:

#include <stdexcept>
long long unsigned log_10(long long unsigned x)
{
     if (x == 0) throw std::out_of_range("Attempted to compute log of zero");
     long long unsigned result_rounded_down = 0;
     long long unsigned power = 10;
     while (power < x)
     {
          ++result_rounded_down;
          power *= 10;
     }
     return result_rounded_down;
}