Mathematical Error

**toonlover** · 11-25-2006

Hello you guys! I don't know why my compiler is giving me this error:

Code:

code.cpp(23) : error C2296: '^' : illegal, left operand has type 'float'
code.cpp(26) : error C2296: '^' : illegal, left operand has type 'float'

And here is the place where I get the error:

Code:

float Height(float G,float V, float S,int POSNEG)
{
      if(POSNEG == POSITIVE)
      {return(((-2*V)/(2*G)) + sqrt((-2*S)/(G)) + ((2*V^2)/(4*G))*2*G);}

      if(POSNEG == NEGATIVE)
      {return(((-2*V)/(2*G)) - sqrt((-2*S)/(G)) + ((2*V^2)/(4*G))*2*G);}

      else
      {
            return ERROR;
      }
}

**The Brain** · 11-25-2006

you are using ^ as an exponentential operation.. when in actuality, it is a bitwise operation using the XOR bitwise operator. common mistake, but easy to fix. try using the pow( ) function from <cmath>

**toonlover** · 11-25-2006

So how exactly do I implement that function in this case??

**The Brain** · 11-25-2006

Code:

((2*V^2)/(4*G))*2*G);

could be written as:

Code:

((pow(2*V, 2)/(4*G))*2*G);

**toonlover** · 11-25-2006

thanx!!

**robatino** · 11-25-2006

Actually, that would be 2*pow(V, 2), assuming that what's meant is 2*(V^2) and not (2*V)^2. In this case it would be easier and more efficient to just write it as 2*V*V.

Edit: And in the second case, 4*V*V.

**grumpy** · 11-25-2006

Originally Posted by robatino

Actually, that would be 2*pow(V, 2), assuming that what's meant is 2*(V^2) and not (2*V)^2.

Indeed. exponentiation has a higher precedence mathematically than multiplication. So (for ^ being exponentiation and * being multiplication) any mathematician would read 2*v^2 as 2*(v^2) rather than as (2*v)^2.

Originally Posted by robatino

In this case it would be easier and more efficient to just write it as 2*V*V.

Indeed. This is one of the simple common tricks used in numerical code. Exponentiation is often markedly less efficient (in terms of machine clock cycles, and therefore runtime speed) than multiplication --- and a compiler has more chance of optimising 2*v*v than it does of optimising 2*pow(v, 2.0).

I would also do some basic algebra (elimination of common terms, precedence of operations) to turn this;

Code:

return (((-2*V)/(2*G)) + sqrt((-2*S)/(G)) + ((2*V*V)/(4*G))*2*G);

into

Code:

   return (sqrt(-2*S/G) + V*V - V/G );

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