How to use log2() function to calculate binary base - 2 logarithm in C++

1 Answer

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#include <iostream>
#include <cmath>

using namespace std;

int main()
{
	cout << "log2(1) = " << log2(1) << endl;
	cout << "log2(1024) = " << log2(1024) << endl;
	cout << "log2(512) = " << log2(512) << endl;
	cout << "log2(5) = " << log2(5) << endl;
	cout << "log2(0) = " << log2(0) << endl;
	cout << "log2(-0) = " << log2(-0) << endl;
	cout << "log2(100) = " << log2(100) << endl;
	cout << "log2(125) = " << log2(125) << endl;
	cout << "log2(125) / log2(5) = " << log2(125) / log2(5) << endl;
	cout << "log2(3.5) = " << log2(3.5) << endl;
	cout << "log2(1000) = " << log2(1000) << endl;
	cout << "log2(0.001) = " << log2(0.001) << endl;

	return 0;

}

/*
run:

log2(1) = 0
log2(1024) = 10
log2(512) = 9
log2(5) = 2.32193
log2(0) = -inf
log2(-0) = -inf
log2(100) = 6.64386
log2(125) = 6.96578
log2(125) / log2(5) = 3
log2(3.5) = 1.80735
log2(1000) = 9.96578
log2(0.001) = -9.96578

*/

 



answered Mar 26, 2016 by avibootz
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