## digitalmars.D.learn - log2 buggy or is a real thing?

- Jonathan M Davis (31/31) Apr 04 2012 This progam:...
- bearophile (36/75) Apr 04 2012 What is the problem you are pe...
- Don Clugston (7/82) Apr 04 2012 I don't think so. For 80-bit r...
- Timon Gehr (4/8) Apr 04 2012 ?...
- Don Clugston (5/15) Apr 04 2012 Ah, you're right. I forgot abo...
- bearophile (12/12) Apr 04 2012 Do you know why is this progra...
- Timon Gehr (2/14) Apr 04 2012 Works correctly here. (x64 lin...
- Don Clugston (2/14) Apr 04 2012 Poor float->decimal conversion...

This progam: import std.math; import std.stdio; import std.typetuple; ulong log2(ulong n) { return n == 1 ? 0 : 1 + log2(n / 2); } void print(ulong value) { writefln("%s: %s %s", value, log2(value), std.math.log2(value)); } void main() { foreach(T; TypeTuple!(byte, ubyte, short, ushort, int, uint, long, ulong)) print(T.max); } prints out 127: 6 6.98868 255: 7 7.99435 32767: 14 15 65535: 15 16 2147483647: 30 31 4294967295: 31 32 9223372036854775807: 62 63 18446744073709551615: 63 64 So, the question is: Is std.math.log2 buggy, or is it just an issue with the fact that std.math.log2 is using reals, or am I completely misunderstanding something here? - Jonathan M Davis

Apr 04 2012

Jonathan M Davis:This progam: import std.math; import std.stdio; import std.typetuple; ulong log2(ulong n) { return n == 1 ? 0 : 1 + log2(n / 2); } void print(ulong value) { writefln("%s: %s %s", value, log2(value), std.math.log2(value)); } void main() { foreach(T; TypeTuple!(byte, ubyte, short, ushort, int, uint, long, ulong)) print(T.max); } prints out 127: 6 6.98868 255: 7 7.99435 32767: 14 15 65535: 15 16 2147483647: 30 31 4294967295: 31 32 9223372036854775807: 62 63 18446744073709551615: 63 64 So, the question is: Is std.math.log2 buggy, or is it just an issue with the fact that std.math.log2 is using reals, or am I completely misunderstanding something here?

What is the problem you are perceiving? The values you see are badly rounded, this is why I said that the Python floating point printing function is better than the D one :-) I print the third result with more decimal digits using %1.20f, to avoid that bad rounding: import std.stdio, std.math, std.typetuple; ulong mlog2(in ulong n) pure nothrow { return (n == 1) ? 0 : (1 + mlog2(n / 2)); } void print(in ulong x) { writefln("%s: %s %1.20f", x, mlog2(x), std.math.log2(x)); } void main() { foreach (T; TypeTuple!(byte, ubyte, short, ushort, int, uint, long, ulong)) print(T.max); print(9223372036854775808UL); real r = 9223372036854775808UL; writefln("%1.20f", r); } The output is: 127: 6 6.98868468677216585300 255: 7 7.99435343685885793770 32767: 14 14.99995597176955822200 65535: 15 15.99997798605273604400 2147483647: 30 30.99999999932819277100 4294967295: 31 31.99999999966409638400 9223372036854775807: 62 63.00000000000000000100 18446744073709551615: 63 64.00000000000000000000 9223372036854775808: 63 63.00000000000000000100 9223372036854775807.80000000000000000000 And it seems essentially correct, the only significant (but very little) problem I am seeing is log2(9223372036854775807) that returns a value > 63, while of course in truth it's strictly less than 63 (found with Wolfram Alpha): 62.999999999999999999843582690243412222355489972678233 The cause of that small error is the limited integer representation precision of reals. In Python there is a routine to compute precisely the logarithm of large integer numbers. It will be useful to have in D too, for BigInts too. Bye, bearophile

Apr 04 2012

On 04/04/12 13:40, bearophile wrote:Jonathan M Davis:This progam: import std.math; import std.stdio; import std.typetuple; ulong log2(ulong n) { return n == 1 ? 0 : 1 + log2(n / 2); } void print(ulong value) { writefln("%s: %s %s", value, log2(value), std.math.log2(value)); } void main() { foreach(T; TypeTuple!(byte, ubyte, short, ushort, int, uint, long, ulong)) print(T.max); } prints out 127: 6 6.98868 255: 7 7.99435 32767: 14 15 65535: 15 16 2147483647: 30 31 4294967295: 31 32 9223372036854775807: 62 63 18446744073709551615: 63 64 So, the question is: Is std.math.log2 buggy, or is it just an issue with the fact that std.math.log2 is using reals, or am I completely misunderstanding something here?

What is the problem you are perceiving? The values you see are badly rounded, this is why I said that the Python floating point printing function is better than the D one :-) I print the third result with more decimal digits using %1.20f, to avoid that bad rounding: import std.stdio, std.math, std.typetuple; ulong mlog2(in ulong n) pure nothrow { return (n == 1) ? 0 : (1 + mlog2(n / 2)); } void print(in ulong x) { writefln("%s: %s %1.20f", x, mlog2(x), std.math.log2(x)); } void main() { foreach (T; TypeTuple!(byte, ubyte, short, ushort, int, uint, long, ulong)) print(T.max); print(9223372036854775808UL); real r = 9223372036854775808UL; writefln("%1.20f", r); } The output is: 127: 6 6.98868468677216585300 255: 7 7.99435343685885793770 32767: 14 14.99995597176955822200 65535: 15 15.99997798605273604400 2147483647: 30 30.99999999932819277100 4294967295: 31 31.99999999966409638400 9223372036854775807: 62 63.00000000000000000100 18446744073709551615: 63 64.00000000000000000000 9223372036854775808: 63 63.00000000000000000100 9223372036854775807.80000000000000000000 And it seems essentially correct, the only significant (but very little) problem I am seeing is log2(9223372036854775807) that returns a value> 63, while of course in truth it's strictly less than 63 (found with Wolfram Alpha): 62.999999999999999999843582690243412222355489972678233 The cause of that small error is the limited integer representation precision of reals.

I don't think so. For 80-bit reals, every long can be represented exactly in an 80 bit real, as can every ulong from 0 up to and including ulong.max - 1. The only non-representable built-in integer is ulong.max, which (depending on rounding mode) gets rounded up to ulong.max+1. The decimal floating point output for DMC, which DMD uses, is not very accurate. I suspect the value is actually <= 63.In Python there is a routine to compute precisely the logarithm of large integer numbers. It will be useful to have in D too, for BigInts too. Bye, bearophile

Apr 04 2012

On 04/04/2012 05:15 PM, Don Clugston wrote:I don't think so. For 80-bit reals, every long can be represented exactly in an 80 bit real, as can every ulong from 0 up to and including ulong.max - 1. The only non-representable built-in integer is ulong.max, which (depending on rounding mode) gets rounded up to ulong.max+1.

? assert(0xffffffffffffffffp0L == ulong.max); assert(0xfffffffffffffffep0L == ulong.max-1);

Apr 04 2012

On 04/04/12 18:53, Timon Gehr wrote:On 04/04/2012 05:15 PM, Don Clugston wrote:I don't think so. For 80-bit reals, every long can be represented exactly in an 80 bit real, as can every ulong from 0 up to and including ulong.max - 1. The only non-representable built-in integer is ulong.max, which (depending on rounding mode) gets rounded up to ulong.max+1.

? assert(0xffffffffffffffffp0L == ulong.max); assert(0xfffffffffffffffep0L == ulong.max-1);

Ah, you're right. I forgot about the implicit bit. 80 bit reals are equivalent to 65bit signed integers. It's ulong.max+2 which is the smallest unrepresentable integer. Conclusion: you cannot blame ANYTHING on the limited precision of reals.

Apr 04 2012

Do you know why is this program: import std.stdio; void main() { real r = 9223372036854775808UL; writefln("%1.19f", r); } Printing: 9223372036854775807.8000000000000000000 Instead of this? 9223372036854775808.0000000000000000000 Bye, bearophile

Apr 04 2012

On 04/04/2012 01:46 PM, bearophile wrote:Do you know why is this program: import std.stdio; void main() { real r = 9223372036854775808UL; writefln("%1.19f", r); } Printing: 9223372036854775807.8000000000000000000 Instead of this? 9223372036854775808.0000000000000000000 Bye, bearophile

Works correctly here. (x64 linux)

Apr 04 2012

On 04/04/12 13:46, bearophile wrote:Do you know why is this program: import std.stdio; void main() { real r = 9223372036854775808UL; writefln("%1.19f", r); } Printing: 9223372036854775807.8000000000000000000 Instead of this? 9223372036854775808.0000000000000000000 Bye, bearophile

Poor float->decimal conversion in the C library ftoa() function.

Apr 04 2012