CODEWITHSUNDEEP
c++
Rittam Das
Rittam Das

Reputation: 9

In C++, log2() and log2l() returning different values

While I was coding in C++ I stumbled upon something, the return value of log2(n) and log2l(n) are completely different for a certain value of n :

#include<bits/stdc++.h>
using namespace std;
int main(){
    int d= log2l(288230376151711743);
    cout<<d;
    return 0;
}

This above code displays a value of 57 whereas the following code:

#include<bits/stdc++.h>
using namespace std;
int main(){
    int d= log2(288230376151711743);
    cout<<d;
    return 0;
}

Displays a value of 58 upon execution.

Why is this occurring can someone please explain me ?

Upvotes: 0

Views: 219

Answers (2)

Rohan Bari
Rohan Bari

Reputation: 7726

Try using the following expression for do-it-yourself:

auto log2 = log2(288230376151711743); // log2 => double | log2l() => long double

You'll get to know that variable d is a datatype set to double, again if you do the same thing with log2l(), you'll find that d is set to long double, i.e. the function log2l() returns a long double, but the value is rounding down while typecasting implicitly from a long double into int.

That's why, it'll show 58 for log2(). And in contrary, 57 for log2l() when the variable is an integer.

Upvotes: 1

Sanjeev Sharma
Sanjeev Sharma

Reputation: 557

It's because you are typecasting double to integer datatype which is rounding it down.

#include <bits/stdc++.h>
using namespace std;

int main() {

    cout << log2(288230376151711743) << "\n";
    cout << log2l(288230376151711743) << "\n";

    cout << (int) log2(288230376151711743) << "\n";
    cout << (int) log2l(288230376151711743) << "\n";

    cout << (long) log2(288230376151711743) << "\n";
    cout << (long) log2l(288230376151711743) << "\n";

    cout << (double) log2(288230376151711743) << "\n";
    cout << (double) log2l(288230376151711743) << "\n";

    return 0;
}

try to run this code and observe the results. log2 docs

Upvotes: 1

Related Questions