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

[ra1u]

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

[tim-one]

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

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

[sunmy2019]

Actually

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

[sunmy2019]

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

[ra1u]

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

[ra1u]

@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))

[sunmy2019]

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__().

[skirpichev]

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.)

[mdickinson]

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