pow is usually evaluated by this formula: x^y = exp2(y*log2(x)) Functions exp2(x),log2(x) are directly implemented in FPU. If you want to implement bignums then they can also be evaluated by basic operators with use of precomputed table of sqrt-powers like: 2^1/2, 2^1/4, 2^1/8, 2^1/16, 2^1/32 … to speed up the process In case you need … Read more
Using the power operator ** will be faster as it won’t have the overhead of a function call. You can see this if you disassemble the Python code: >>> dis.dis(‘7. ** i’) 1 0 LOAD_CONST 0 (7.0) 3 LOAD_NAME 0 (i) 6 BINARY_POWER 7 RETURN_VALUE >>> dis.dis(‘pow(7., i)’) 1 0 LOAD_NAME 0 (pow) 3 LOAD_CONST … Read more
As of C++11, special cases were added to the suite of power functions (and others). C++11 [c.math] /11 states, after listing all the float/double/long double overloads (my emphasis, and paraphrased): Moreover, there shall be additional overloads sufficient to ensure that, if any argument corresponding to a double parameter has type double or an integer type, … Read more
You can write your own, using repeated squaring: BigInteger pow(BigInteger base, BigInteger exponent) { BigInteger result = BigInteger.ONE; while (exponent.signum() > 0) { if (exponent.testBit(0)) result = result.multiply(base); base = base.multiply(base); exponent = exponent.shiftRight(1); } return result; } might not work for negative bases or exponents.
The solution for arguments under 1.7976931348623157E308 (Double.MAX_VALUE) but supporting results with MILLIONS of digits: Since double supports numbers up to MAX_VALUE (for example, 100! in double looks like this: 9.332621544394415E157), there is no problem to use BigDecimal.doubleValue(). But you shouldn’t just do Math.pow(double, double) because if the result is bigger than MAX_VALUE you will just … Read more
pow() returns numbers as floating point. What is really returning is 24.99997 or something similar. But then you truncate it by assigning it to an integer and it comes out as 24. It’s ok to assignit to an integer in this case because you know then answer will always be an integer so you need … Read more
Add 0.5 before casting to int. If your system supports it, you can call the C99 round() function, but I prefer to avoid it for portability reasons.
Due to the representation of floating point values pow(10.0, 5) could be 9999.9999999 or something like this. When you assign that to an integer that got truncated. EDIT: In case of cout << pow(10.0, 5); it looks like the output is rounded, but I don’t have any supporting document right now confirming that. EDIT 2: … Read more
MethodImplOptions.InternalCall That means that the method is actually implemented in the CLR, written in C++. The just-in-time compiler consults a table with internally implemented methods and compiles the call to the C++ function directly. Having a look at the code requires the source code for the CLR. You can get that from the SSCLI20 distribution. … Read more