How do I calculate power-of in C#?

Question

I\'m not that great with maths and C# doesn\'t seem to provide a power-of function so I was wondering if anyone knows how I would run a calculation like this:

Answer 1

Math.Pow() returns double so nice would be to write like this:

double d = Math.Pow(100.00, 3.00);

Answer 2

Following is the code calculating power of decimal value for RaiseToPower for both -ve and +ve values.

public decimal Power(decimal number, decimal raiseToPower)
        {
            decimal result = 0;
            if (raiseToPower < 0)
            {
                raiseToPower *= -1;
                result = 1 / number;
                for (int i = 1; i < raiseToPower; i++)
                {
                    result /= number;
                }
            }
            else
            {
                result = number;
                for (int i = 0; i <= raiseToPower; i++)
                {
                    result *= number;
                }
            }
            return result;
        }

Answer 3

I'm answering this question because I don't find any answer why the ^ don't work for powers. The ^ operator in C# is an exclusive or operator shorten as XOR. The truth table of A ^ B shows that it outputs true whenever the inputs differ:

Input A	Input B	Output
0	0	0
0	1	1
1	0	1
1	1	0

Witch means that 100 ^ 3 does this calculation under the hood:

hex   binary
100 = 0110 0100 
  3 = 0000 0011
---   --------- ^
103 = 0110 0111

With is of course not the same as Math.Pow(100, 3) with results to one million. You can use that or one of the other existing answers.

---

You could also shorten your code to this that gives the same result because C# respects the order of operations.

double dimensions = 100 * 100 / Math.Pow(100, 3); // == 0.01

Answer 4

Do not use Math.Pow

When i use

for (int i = 0; i < 10e7; i++)
{
    var x3 = x * x * x;
    var y3 = y * y * y;
}

It only takes 230 ms whereas the following takes incredible 7050 ms:

for (int i = 0; i < 10e7; i++)
{
    var x3 = Math.Pow(x, 3);
    var y3 = Math.Pow(x, 3);
}

Answer 5

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

Answer 6

You are looking for the static method Math.Pow().