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What does ^ operator do?

开发者 https://www.devze.com 2023-03-16 15:56 出处:网络
I thought that ^ did that. I expected: 10^0=1 10^1=10 10^2=100 What I\'m getting 10^0=10 10^1=11 10^2=8 the actual code is

I thought that ^ did that. I expected:

10^0=1
10^1=10
10^2=100

What I'm getting

10^0=10
10^1=11
10^2=8

the actual code is

int value = 10 ^ exp;

repl开发者_Python百科acing exp for 0, 1, and 2 What does the ^ operator do?


Math.Pow(x, y) to get x raised to the power of y. You were doing an XOR. C# operators


You want to do:

Math.Pow(10, exp);

This actually produces a double though so you'll need to cast it down if you really want an int.


In c#, ^ is the logical XOR operator.


As others have said, you need you use Math.pow(x, y) to do x^y in C#. The ^ operator is actually a logical XOR operator on the bits of the two numbers. More information on the logical XOR can be found here: http://msdn.microsoft.com/en-us/library/zkacc7k1.aspx


Math.Power(10,exp) works like charm...


In C#, the ^ operator is the logical XOR operator.

http://msdn.microsoft.com/en-us/library/6a71f45d.aspx


^ is the XOR operator.

http://en.wikipedia.org/wiki/Xor

If you want to raise a number to a power, use

Math.Pow(num, exp);


There is no operator for this. Use Math.Pow.

There is a full list of all the C# operators with linked documentation on each one.


The thing is that ^ is equivalent for XOR, that's why 10 ^ 0 = 10 because

1010 XOR    1010 XOR
0000 =      0010 =
1010        1000

You need to use Math.Pow method.

P.S. = 10^2 actually returns 8...

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