为什么“％.10f”可以％Decimal（u）发出带有文字冒号的字符串？ [英] Why can "%.10f" % Decimal(u) emit a string with a literal colon?
问题描述
格式化要打印的数字时，点后紧接着用冒号格式化12位数字。为什么会这样呢？这是AIX系统上的Python 2.7。
When formatting a number to be printed, 12 digit numbers are being formatted with a colon immediately after the dot. Why is this happening? This is Python 2.7 on an AIX system.
$ uname -a ; /opt/bin/python2.7
AIX myserver 1 6 00F6A5CC4C00
Python 2.7.12 (default, Sep 29 2016, 12:02:17) [C] on aix5
Type "help", "copyright", "credits" or "license" for more information.
>>> '{0:.10f}'.format(123456789012)
'123456789011.:000000000'
>>> from decimal import Decimal
>>> u=123456789012
>>> print "%.10f" % Decimal(u)
123456789011.:000000000
其他信息：
不是每12位数字：
>>> for x in range(123456789010,123456789020):
... print '{0:.10f}'.format(x)
...
12345678900:.0000000000
123456789010.:000000000
123456789011.:000000000
123456789013.0000000000
123456789013.:000000000
123456789015.0000000000
123456789016.0000000000
123456789017.0000000000
123456789017.:000000000
123456789019.0000000000
其他任何长度的数字都不会发生这种情况。另外，我尝试了bash和perl的printf，但是这两个都没有发生。
This is not happening with any other length numbers. Also, I tried bash's and perl's printf and this is not happening with either of them.
这里发生了什么？
根据要求，这是屏幕截图视频。
更多请求的信息：
>>> import locale
>>> locale.getdefaultlocale()
('en_US', 'ISO8859-1')
结果的user2357112粘贴代码：
Result of user2357112 pastebin code:
>>> import ctypes
>>> f=ctypes.pythonapi.PyOS_double_to_string
>>> f.argtypes=ctypes.c_double,ctypes.c_char,ctypes.c_int,ctypes.c_int,ctypes.POINTER(ctypes.c_int))
>>> f.restype=ctypes.c_char_p
>>> print f(123456789012.0, 'f', 10, 0, None)
123456789011.:000000000
Antti_Happa的pastebin正确打印了所有数字。
Antti_Happa's pastebin printed all numbers correctly.
使用格式r给出：
print 'e: {0:.10e}\nf: {0:.10f}\ng: {0:.10g}\nr: {0:0r}'.format(x)
ValueError: Unknown format code 'r' for object of type 'int'
使用e，f和g格式提供以下内容：
Using e, f and g formats provided the following:
for x in range(123456789000,123456789019):
print 'e: {0:.10e}\nf: {0:.10f}\ng: {0:.10g}'.format(x)
e: 1.2345678900e+11
f: 123456789000.0000000000
g: 1.23456789e+11
e: 1.2345678900e+11
f: 123456789000.:000000000
g: 1.23456789e+11
e: 1.2345678900e+11
f: 123456789001.:000000000
g: 1.23456789e+11
e: 1.2345678900e+11
f: 123456789003.0000000000
g: 1.23456789e+11
我无权在此服务器上安装或更新任何内容。我可以请求更新的版本，但是这种性质的更改请求需要花费大量时间。另外，其他程序也依赖于此安装，并且需要进行大量测试。
I have no access to install or update anything on this server. I can request an updated version, but change requests of this nature take a fair amount of time. Also, other programs depend on this installation and a lot of testing would be required.
我被告知仅将安装IBM提供的软件包，而最新的python 2.7 IBM提供的软件包是2.7.12。
I have been informed that only IBM provided packages will be installed and that the latest python 2.7 package provided by IBM is 2.7.12.
我已经通过这样做来解决了这个问题
I have "fixed" the problem by doing
othervar = '{0:.10f}'.format(somevar).replace(':', '0')
我知道这是非常不安全的，但是... <耸耸肩
which is extremely unsafe, I know, but ... shrug
啊！我只是注意到一个不合一的错误... 123456789012 被格式化为以下格式： 123456789011.:0000000000 ...这是一个奇怪的错误。
Argh! I just noticed an off-by-one error ... 123456789012 is being formatted as one less: 123456789011.:0000000000 ... this is a weird bug.
推荐答案
虽然不是答案，但我可以在
Although not an "answer", I can give you my results on a slightly newer version running on AIX.
对不起，我无法复制您的问题。
Sorry that I'm unable to replicate your problem.
[lholtscl@ibm3 ~]$ python
Python 2.7.13 (default, Sep 7 2017, 21:08:50) [C] on aix7
Type "help", "copyright", "credits" or "license" for more information.
>>> print "%.10f" % 123456789012
123456789012.0000000000
>>> '{0:.10f}'.format(123456789012)
'123456789012.0000000000'
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