Is java.lang.Math.PI equal to GCC's M_PI?

Question

I am coding several reference algorithms in both Java and C/C++. Some of these algorithms use π. I would like for the two implementations of each algorithm to produce identical results, without rounding differently. One way to do this that has worked consistently so far is to use a custom-defined pi constant which is exactly the same in both languages, such as 3.14159. However, it strikes me as silly to define pi when there are already high-precision constants defined in both the Java and GCC libraries.

I've spent some time writing quick test programs, looking at documentation for each library, and reading up on floating-point types. But I haven't been able to convince myself that java.lang.Math.PI (or java.lang.StrictMath.PI) is, or is not, equal to M_PI in math.h.

GCC 3.4.4 (cygwin) math.h contains:

#define M_PI            3.14159265358979323846
                                         ^^^^^

but this

printf("%.20f", M_PI);

produces

3.14159265358979311600
                 ^^^^^

which suggests that the last 5 digits cannot be trusted.

Meanwhile, Javadocs say that java.lang.Math.PI is:

The double value that is closer than any other to pi, the ratio of the circumference of a circle to its diameter.

and

public static final double PI  3.141592653589793d

which omits the questionable last five digits from the constant.

System.out.printf("%.20f\n", Math.PI);

produces

3.14159265358979300000
                 ^^^^^

If you have some expertise in floating-point data types, can you convince me that these library constants are exactly equal? Or that they are definitely not equal?

Answer 1

Note the following.

The two numbers are the same to 16 decimal places. That's almost 48 bits which are the same.

In an IEEE 64-bit floating-point number, that's all the bits there are that aren't signs or exponents.

The #define M_PI has 21 digits; that's about 63 bits of precision, which is good for an IEEE 80-bit floating-point value.

What I think you're seeing is ordinary truncation of the bits in the M_PI value.

Answer 2

What you want to do is print out the raw bit pattern for the PI values and compare them.

In Java use the http://java.sun.com/j2se/1.5.0/docs/api/java/lang/Double.html#doubleToRawLongBits(double) method to get the long value that you should print as binary.

Java 5 gives:

• PI is 3.141592653589793
• Raw bits is 4614256656552045848
• Binary is 100000000001001001000011111101101010100010001000010110100011000

In C, you can do double pi = M_PI; printf("%lld\n", pi); to obtain the same 64bit integer: 4614256656552045848 (thanks Bruno).