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Comment by CalChris | Hacker News Reader

  #define M_PI        3.14159265358979323846264338327950288

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3.141592653589793115997963468544185161590576171875 is the exact decimal representation for the 64-bit IEEE-754 number that's closest to pi (viz. 0x400921FB54442D18).

Any time you see a decimal floating-point constant with a nonzero fractional part that doesn't end in '5', you're looking at a bug.

EDIT: As long as this grizzled old Fortran programmer is giving out free advice, I'll add two more items every programmer should know about binary floating-point:

a) Every binary floating-point number can be represented exactly in decimal notation if you use enough digits.

b) Those decimal values are the only ones that can be exactly converted to binary; all of the rest require rounding.

> Any time you see a decimal floating-point constant with a nonzero fractional part that doesn't end in '5', you're looking at a bug.

That's just silly. If you're writing some famous mathematical constant, the digits should match that constant, and not the requirements of the machine. (Except for the last one being rounded off.) Suppose we had a floating-point machine that gave us maximum 4 digits of decimal precision. I wouldn't define the PI constant as 3.145. That would just look like a typo to people who have PI memorized to half a dozen digits or more. I'd make it 3.14159 (or more) and let the darn compiler find the nearest approximation on the floating-point axis.

Any exact decimal representation of a specific binary floating-point number that's finite and not an integer must end in the digit '5' (perhaps with trailing zeroes). This is because its fractional part is (the sum of) a set of powers of two with negative exponents, and their exact decimal representations (0.5, 0.25, 0.125, &c.) all end in '5' (proof by induction is obvious and left to the reader).

Any time you see a decimal floating-point constant with a nonzero fractional part that doesn't end in '5', you're looking at a bug.

depends on the language. for example here it is in go:

    Pi  = 3.14159265358979323846264338327950288419716939937510582097494459 // http://oeis.org/A000796

What the person above you is saying, I think, is to remember that computers usually work in base 2. This applies to IEEE floating points, where the mantissa is in base 2; when you represent fractions in base two, they're powers of two: 1/2 (.5), 1/4 (.25), 1/8 (.125), etc. What he's asserting, I think, is that any power-of-two fraction, or any combination of those (in binary), result in a number ending in 5 when represented in decimal. Anything else is going to be rounded to the nearest representable number (that ends in 5).

So, go might have that value in its source, but it's getting rounded to something that would, if represented in decimal, end in 5.

In Go, floating-point constants may have very high precision, so that arithmetic involving them is more accurate. The constants defined in the math package are given with many more digits than are available in a float64.

Having so many digits available means that calculations like Pi/2 or other more intricate evaluations can carry more precision until the result is assigned, making calculations involving constants easier to write without losing precision. It also means that there is no occasion in which the floating-point corner cases like infinities, soft underflows, and NaNs arise in constant expressions.

There's an argument to be made the other way too. If you're using an unusual default rounding direction and you care which direction your PI constant is rounded, you might prefer it if PI rounded in the same rounding direction as the rest of your floating point math. In that case you'd want a constant that is equivalent to PI under all rounding modes.

The code below works for me

#define PI1 3.141592653589793115997963468544185161590576171875

double pi1 = PI1;

#define PI2 3.14159265358979323846264338327950288

double pi2 = PI2;

assert(pi1 == pi2);

(edit: or even 3.14159265358979323846)

Your PI2 rounds to PI1 under the rounding mode used at compilation time. Print it out with "%50.48f" (or FORMAT(F50.48)) and you'll see PI1. But PI1 is independent of rounding mode.

Sure. I thought you were implying more than that. When you called it a bug I thought you were implying that the use of one over another would alter the output of a program. My bad.

I thought that style was considered bad form in modern C/C++ (esp. the later, with `constexpr`). What's wrong with

    const PI: f64 = 3.14159265358979323846264338327950288;

?

What machine can represent all those decimal digits so precisely that the ending decimal digits ...0288 are exactly right and not ...0287 ?

I'll accept without looking it up that the statement is correct syntax in some language (it doesn't look like C++ to me).

See above. The constant is (a) not pi, and (b) not an exact decimal representation of any binary floating-point number.

const M_PI: f64 = 3.14159265358979323846264338327950288;

You wouldn't use a macro for something like that in Rust.

I don't understand what this means. You'd use const in Rust for this, not a macro.

That was just a trivial example. Yes, you could use a const which would be difficult to share across files. Moreover, if you wanted to do something equally textual

  #define PASS_VERBOSE if (flag_verbose && first_pass) printf

you just might be able to but only with a completely different tool.

A macro processor is not of the language; it is above the language.

A "macro" in C is not the same thing as a "macro" in Rust. By your logic, Erlang processes aren't processes because they aren't kernel processes, or Go packages aren't packages because they don't use the Java package naming convention. Nobody owns the exclusive right to define the words we use.

> A macro processor is not of the language; it is above the language.

Then feel free to use the C preprocessor with Rust, it works just as well. :P Just like it does with Python, and Java, and...

The C preprocessor can only be used with Python as long as you don't do anything multi-line:

   #define whatever(param) \
   foo: \
      bar \
      baz

I made a preprocessor some 18 years ago that could be used with Python.

Wayback Machine:

https://web.archive.org/web/20000815202258/http://users.foot...

Better yet, use m4.

> which would be difficult to share across files.

It's path::to::wherever::PI. That's it. Just like any other item. If you want to use only PI in your code, you'd use 'use', like any other name.

Your second example is something better suited to a macro, it's true.

I _think_ what you're getting at here is that you only want text substitution? I think we will have to agree to disagree if that's true :)

Well, given that you have only written two macros in your years of Rust, I would strongly encourage you or the language ergonomics initiative to openly question why this is so. Clearly, Rust has a clever approach but I'm questioning whether it is in fact a usable approach. Too much solution for not enough problem.

Yes, I do like textual substitution. Guilty. This is a common old school low level paradigm. Still, the underlying language and its compiler exist below to enforce the rules on any atrocities I commit with my macros.

If you want textual substitution, you are not limited by anything that Rust offers. You can incorporate any existing text preprocessor - of which there's a multitude, and at least cpp and m4 are in pretty much any Unix system - into your compile pilelines. By the very definition of textual substitution, it is completely orthogonal to the meaning of the output (in this case, Rust code).

The kind of macros Rust implements, on the other hand, are the kind that have to be language-specific, because they deal with the syntax tree of that particular language. They also enable many things that are outright impossible with text substitution, such as hygienic macros.

> Clearly, Rust has a clever approach but I'm questioning whether it is in fact a usable approach. [...] Yes, I do like textual substitution.

These two statements are not consistent. Textual substitution is not usable. The pitfalls with it have been well documented for decades, and yet the same problems persist. Why on earth would you want to perpetuate the list of problems caused by such a facility?

Syntactic macros are absolutely superior to textual substitution. You can take issue with Rust's currently limited support for macros, but to propose textual substitution as a viable and more usable alternative is simply absurd.

What I'm saying is, I personally find Rust expressive enough to never need macros, and so their ease of use, to me personally, is not an issue either way. (Well, other than the import bit, I do care about that.)

Those who do use and write them heavily are the ones actually involved in the macros 2.0 effort. There are certainly flaws.

You know what I like? Sensible error messages, which you tend to lose if you use macros for anything complex (disclaimer: I am very guilty here)

I very rarely write macros despite having written thousands of lines of Rust, but they are super useful on occasion. Often in Rust something that you might have used preprocessor macro tricks to do in C can be done without macros at all. At other times, Rust macros will let you do things that would be impossible to express in C.

You want textual macros. Rust doesn't have this feature, it has a syntactic macro feature inspired by the Lisp family. Sorry.

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