Is There an Exponent Operator in C#

Is there an exponent operator in C#?

The C# language doesn't have a power operator. However, the .NET Framework offers the Math.Pow method:

Returns a specified number raised to the specified power.

So your example would look like this:

float Result, Number1, Number2;

Number1 = 2;
Number2 = 2;

Result = Math.Pow(Number1, Number2);

Exponent operator

The e is not an operator. Is is part of the real literal syntax of the C# language. It is evaluated when the code is parsed and can only be used to express a constant literal.

Just like you can use suffixes f to denote a float, or d to denote a double, the XeY syntax allows you to define doubles with the value X * 10^Y. It’s simply meant as a convenience when you’re writing code with constant numbers.

Here are some more examples on what you can do with all these different syntaxes, and which type and constant literal they are equivalent to:

var a = 2e5;      // (double) 200000
var b = 2.3e3;    // (double) 2300
var c = 13f;      // (single) 13
var d = 53d;      // (double) 53
var e = 1.2e3f;   // (single) 1200
var f = 2.3m;     // (decimal) 2.3
var g = 1.23e-4m; // (decimal) 0.000123

However, none of that is exposed to the .NET runtime, it is all evaluated at compile time and stored as exposed as constant literals—there is absolutely no difference between the short syntax and the full constant. As such, this functionality is not available for variables which only exist at run-time. If you need to exponentiate those, you will have to use other methods, for example myVariable * 10e5.

How to do exponentiation in constant expression?

Since in your particular case you want to raise 2 into MaxExponent power

2 ** MaxExponent

you can put it as a left shift, but if and only if MaxExponent is a small positive integer value:

1 << MaxExponent

Like this

// double: see comments below `1L` stands for `long` and so MaxExponent = [0..63]   
public const double MaxValue = MaxMantissa * (1L << MaxExponent);

In general case (when MaxExponent is an arbitrary double value), you can try changing const to readonly

public static readonly double MaxValue = MaxMantissa * Math.Pow(2.0, MaxExponent);

Overloaded exponents C# and exponent operator

You could create an extension method to act as a shortcut to Math.Pow(), this is actually a vey common thing to do.

perhaps:

public static double Pow(this double a, double b)
{
    return Math.Pow(a, b);
}

So then you can use it like this:

myDouble = myDouble.Pow(3);

There's various formats you can play around with when it comes to extension methods. If your goal is brevity, I'm sure you can come up with something you'll be happy with.

Math.Pow() vs Math.Exp() C# .Net

Math.Pow computes x ^y for some x and y.

Math.Exp computes e ^x for some x, where e is Euler's number.

Note that while Math.Pow(Math.E, d) produces the same result as Math.Exp(d), a quick benchmark comparison shows that Math.Exp actually executes about twice as fast as Math.Pow:

Trial Operations       Pow       Exp
    1       1000 0.0002037 0.0001344 (seconds)
    2     100000 0.0106623 0.0046347 
    3   10000000 1.0892492 0.4677785 

How do you do *integer* exponentiation in C#?

A pretty fast one might be something like this:

int IntPow(int x, uint pow)
{
    int ret = 1;
    while ( pow != 0 )
    {
        if ( (pow & 1) == 1 )
            ret *= x;
        x *= x;
        pow >>= 1;
    }
    return ret;
}

Note that this does not allow negative powers. I'll leave that as an exercise to you. :)

Added: Oh yes, almost forgot - also add overflow/underflow checking, or you might be in for a few nasty surprises down the road.

C# how to do e.g. 5*10^4 or numbers to the power of ten etc

You can write the literal float constant 5.0e4 (and it works in C, C++, Java, Fortran also).

How do I calculate power-of in C#?

See Math.Pow. The function takes a value and raises it to a specified power:

Math.Pow(100.00, 3.00); // 100.00 ^ 3.00

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