log(Decimal(x)) is broken for large x

issue is that for large x math.log(Decimal(x)) returns inf

I am not sure if that is an issue with documentation or implementation

Python 3.11.4 (main, Jun  7 2023, 10:13:09) [GCC 12.2.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> from math import log
>>> from decimal import Decimal
>>> d = Decimal(300**300)
>>> d
Decimal('136891479058588375991326027382088315966463695625337436471480190078368997177499076593800206155688941388250484440597994042813512732765695774566001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000')
>>> log(d)
inf
>>> log(int(d))
1711.1347423968602

The math module is really for floats. decimal has its own algorithms:

>>> import decimal
>>> x = decimal.Decimal(300**300)
>>> x.ln()
Decimal('1711.134742396860317829368444')

Actually

>>> float(decimal.Decimal(300**300))
inf

You should not rely on math.log which is designed for float.

I understand all of that. What I am asking is following: Does documentation/specification states this restrictions?

@sunmy2019

@sunmy2019

There is another issue with float() implementation .

From: https://docs.python.org/3/library/functions.html?#float

If the argument is outside the range of a Python float, an OverflowError will be raised.

According to this OverflowError should be raised from float(decimal.Decimal(300**300))

If the argument is outside the range of a Python float, an OverflowError will be raised.

That's only true for int and float.

Otherwise, if the argument is an integer or a floating point number, a floating point number with the same value (within Python’s floating point precision) is returned. If the argument is outside the range of a Python float, an OverflowError will be raised.

You should look at the following paragraph for general Python objects.

For a general Python object x, float(x) delegates to x.__float__(). If __float__() is not defined then it falls back to __index__().

@tim-one

On one hand, it might be considered as a tiny documentation issue for the float() type.

Here is a patch.

diff --git a/Doc/library/functions.rst b/Doc/library/functions.rst
index 3520609706..2672d243c7 100644
--- a/Doc/library/functions.rst
+++ b/Doc/library/functions.rst
@@ -677,7 +677,7 @@ are always available.  They are listed here in alphabetical order.
 
    Otherwise, if the argument is an integer or a floating point number, a
    floating point number with the same value (within Python's floating point
-   precision) is returned.  If the argument is outside the range of a Python
+   precision) is returned.  If an integer is outside the range of a Python
    float, an :exc:`OverflowError` will be raised.
 
    For a general Python object ``x``, ``float(x)`` delegates to

On another hand, there could be some confusion in this case:

>>> log(int(d))
1711.1347423968602
>>> float(int(d))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
OverflowError: int too large to convert to float

While for other functions of the module, like sin(), we could expect simple type-cast for integer arguments, i.e. sin(<big-int>)=sin(float(<big-int>)) - the log() function is a special one. It uses internally the _PyLong_Frexp() private helper to compute the log even in some cases when an integer argument is outside the range of a Python float.

@tim-one, shouldn't we document this somehow? Maybe we could expose the _PyLong_Frexp() in the math.frexp() do deal with big integers in this function like in the log()? (This will break the x == m * 2**e invariant, however.)

I think this should be closed. We can open a separate issue for improving the documentation if that's needed.

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log(Decimal(x)) is broken for large x #106502

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log(Decimal(x)) is broken for large x #106502

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tim-one commented on Jul 7, 2023

sunmy2019 commented on Jul 7, 2023

sunmy2019 commented on Jul 7, 2023

ra1u commented on Jul 7, 2023

ra1u commented on Jul 7, 2023

sunmy2019 commented on Jul 7, 2023

skirpichev commented on Oct 4, 2023

mdickinson commented on Jun 24, 2024

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