Python中的倍数与C ++的精度相同

Question

I am rewriting a C++ program into Python. I need to multiply 2 doubles but C++ and Python don't give the same result. Here's an example with 'hard-coded' values:

C++

printf("%f", ( 44474025505478620106407223274000875520.0 * 5454277033526873088.0 ) );
>>> 242573655903020442240866171189072992939998568974355791872.0

Python

print("%f" % ( 44474025505478620106407223274000875520.0 * 5454277033526873088.0 ) )
>>> 242573655903020398684723205308949669628048817708024725504.0

My problem is that I don't need the most accurate result: I need to get (with Python) a result as close as C++'s result as possible.

In my example, the 15 first digits are the same:

C++ > 242573655903020[442240866171189072992939998568974355791872.0
Py  > 242573655903020[398684723205308949669628048817708024725504.0

I need to have a result even more close (18 first digits would be nice)

I'm really stuck here... Anybody has an idea?

FYI :

Python version: 2.7.8

C++ compiler: cl.exe (the one from visual studio)

Answer 1

It seems to depend on the Python implementation. For example, with ideone (cpython 2.7.13), I'll get the same result as your C result.

C version on Ideone - Result:

242573655903020442240866171189072992939998568974355791872.000000

Python version on Ideone - Result:

242573655903020442240866171189072992939998568974355791872.000000

Answer 2

Use library decimal , take your snippet as an example:

from decimal import Decimal

print("%f" % ( Decimal("44474025505478620106407223274000875520.0") * Decimal("5454277033526873088.0") ) )

It gives 242573655903020442240866171189072992939998568974355791872.000000 which is exactly the same as the result given in C .

Answer 3

DBL_DIG or std::numeric_limits::digits10 will probably return 15. With IEE754 doubles you get 15-17 digits depending on the number. Your result is within spec. You can mitigate to achieve higher precision by using multiprecision numbers. In C++ Boost Multiprecision is an option as is something like mpmath in python

http://www.boost.org/doc/libs/1_65_1/libs/multiprecision/doc/html/index.html

http://mpmath.org/

Answer 4

you could use python GMPY library, or you could just use python's "bignumber" . since there is no limit for int in python , you could just do

fl1*10**len(fl1)*fl2*10**len(fl2)

then divide with

10**(len(fl2)+len(fl1))