Power Funtion problem C++

[punjboy]

Hi,

I am using a power funtion in my program and i am getting errors and apparently my program is not accepting the changes.

I am using the 'pow' function. the formulas i am writing are - Diameter = B Volume 1/3 and Pressure = Volume 1/ B

and i am using #include <cmath>


/*************************************…
** Diameter
**************************************…

float balloon::diameter()
{
return (B * (float) pow(Volume,(1/3))); // Calculating Diameter
}


/*************************************…
** Pressure
**************************************…

float balloon:ressure()
{
return ((float) pow(Volume,(1/B))); // Caluclating Pressure
}



the errors i am getting with this is -

\visual studio 2008\projects\prog8\code_2.cpp(71) : warning C4244: 'return' : conversion from 'double' to 'float', possible loss of data
\visual studio 2008\projects\prog8\code_2.cpp(81) : error C2666: 'pow' : 6 overloads have similar conversions
c:\program files\microsoft visual studio 9.0\vc\include\math.h(575): could be 'long double pow(long double,int)'
c:\program files\microsoft visual studio 9.0\vc\include\math.h(573): or 'long double pow(long double,long double)'
1> c:\program files\microsoft visual studio 9.0\vc\include\math.h(527): or 'float pow(float,int)'
1> c:\program files\microsoft visual studio 9.0\vc\include\math.h(525): or 'float pow(float,float)'
1> c:\program files\microsoft visual studio 9.0\vc\include\math.h(489): or 'double pow(double,int)'
1> c:\program files\microsoft visual studio 9.0\vc\include\math.h(123): or 'double pow(double,double)'
1> while trying to match the argument list '(float, double)'
1>Build log was saved at "file://c:\Users\Aman\Documents\Visual Studio 2008\Projects\Prog8\Debug\BuildLog.htm"
1>Prog8 - 1 error(s), 1 warning(s)


Would appreciate any help...

Thanks

[GCDEF]

We don't know what Volume or B are, and apparently the compiler's not sure either.

[punjboy]

I guess i should rephrase the question -

how would u enter this in a c++ format - ( using power function POW)

B *air 1/3

[Paul McKenzie]

I guess i should rephrase the question -

how would u enter this in a c++ format - ( using power function POW)

B *air 1/3

This will answer your own question.

What does this program output:

#include <iostream>

int main()
{
   std::cout << 1/3;
}

Now what does this output?

#include <iostream>

int main()
{
   std::cout << 1.0/3.0;
}

See the difference? What is the difference between 1, 1.0, and 3, 3.0?

Regards,

Paul McKenzie

[punjboy]

Well, i did try that

B * float pow (float Volume,float (1.0/3.0));


but it doesn't gives me the right answer ..

[Paul McKenzie]

Well, i did try that

B * float pow (float Volume,float (1.0/3.0));


but it doesn't gives me the right answer ..

It shouldn't have even compiled. Do you have a C++ book, or are you just trying to make things up and seeing what works? Basically, that line of code doesn't make any sense,

It's very simple:

#include <cmath>
#include <iostream>

int main()
{
   float Volume = 45.0F;
   float Value = pow(Volume, float(1.0/3.0));
   std::cout << Value;
}

If you're using float, then the pow() function requires two float arguments. If you're using double, then the pow() function requires two double arguments (or a double and an int as the power).

Note that I stated 1.0/3.0 and not 1/3. Do you know the difference? It is very important to this entire discussion.

Regards,

Paul McKenzie