Pi输出迅速

Question

I am trying to build a degrees/radians calculator. Here are the functions I have.

func degreesToRadians(degrees: Double) -> Double {

    return degrees * (M_PI / 180)

}

func radiansToDegrees(radians: Double) -> Double {

    return radians * (180 / M_PI)

}

degreesToRadians(90)

radiansToDegrees(M_PI/4)

degreesToRadians(90) returns 1.5707963267949 radiansToDegrees(M_PI/4) returns 45.0

What I want to happen is in the Degrees to Radians function instead of 1.5707963267949 I want the output to be π/2. Is this even possible?

Thanks.

Answer 1

If you want exactly: "π/2", I don't think this is possible with double. You might get this in String, but not in a number. I think your best option would be to take an optional bool in which case result is returned as multiple of pi.

degreesToRadians(90, resultAsMultipleOfPi:true)

If this bool is true then 0.5 should be returned. You may need to do some rounding off to get well rounded numbers.

Answer 2

If you look closely on what is really M_PI you will see that it is predefined double value in math.h that equals

#define M_PI        3.14159265358979323846264338327950288   /* pi             */

If you want more precision you can declare and use long double value instead.

Answer 3

To represent 90 degrees as π/2 , what you want to do is

• consider the fraction of degrees over 180 (ie numerator of degrees and denominator of 180 );

• reduce this fraction (ie divide both the numerator and denominator by the greatest common factor ); in the case of 90/180 , the greatest common factor is, in fact, 90 , yielding a reduced numerator of 1 and a reduced denominator of 2 ;

• the result as a string representation of the fraction, ie something like

   "\\(reducedNumerator)π/\\(reducedDenominator)" 

  Obviously, if either the numerator or denominator are 1 , then you can suppress that portion of the string, but hopefully you get the basic idea.

Take a crack at that.

Answer 4

You can directly use M_PI constant.

If you need in float format just use this,

let pi = Float(M_PI)