C++ math operators

I'm learning C++ and need to know if there is a math operator for raising a number to a power. Example 2 sqaured would be written how???? (Other than 2 * 2)

You can use the power operator: ^

eg. 2 to the power of three would be 2^3;

Thank you

C++ does not have a ^ operator. Use the pow() function.

Example: 2 to the power of 3 would be pow(2,3)

Oops. Sorry My mistake - that's VB, not C++

Well now I have a new problem. I have been trying to figure it out for hours, and no luck.

I'm trying to have the program calculate this equation:
N = P * 2 ^ (T/10). T is days, and integer. P is any whole number, an integer. N should be a decimal (double), but it always comes out to be another whole number.

I'm using a loop to calculate N, when P is entered by the user, for 10 days.

IE:

for (T=1; T<=10; T++)
{
N = P * (2^(T/10));
cout << T;
cout << " " ;
cout << N << '\n';
}

This is probably a very simple thing, but I can't figure out why the output is wrong.

Any ideas?

Ok, I did this once. if you are calculating with integers the result will be integer. try for instance printing the value of you T/10.... you wold be surprised. Even if you know these T and P should be integers, declare them as floats, and see what happens.

Now I have a little prolem: does anybody know what is the function in c++ which gives the maximum between two values?

cheers,
Kazu.

Kazu is essentially right, but I wouldn't declare an integer as a float if it should be an integer. Declare T and P as integers, but convert them for the calculation:

for (T=1; T<=10; T++)
{
N = static_cast<double>P * (2^(static_cast<double>T/10.0));
cout << T;
cout << " " ;
cout << N << '\n';
}

This way you retain the benefits of knowing that T and P are integers, as they are supposed to be for the rest of the program (also notice the division is by 10.0).

Hope this was in time to help...