注意:此问题仅供引用。我很想知道如何深入了解 Python 的内部结构。

不久前，某家内部开始讨论question关于在调用 print 之后/期间是否可以修改传递给打印语句的字符串已经完成。例如，考虑函数:

def print_something():
    print('This cat was scared.')

This dog was scared.

There are a couple of ways to do that, but they're all very ugly, and should never be done. The least ugly way is to probably replace the code object inside the function with one with a different co_consts list. Next is probably reaching into the C API to access the str's internal buffer. [...]

>>> import inspect
>>> exec(inspect.getsource(print_something).replace('cat', 'dog'))
>>> print_something()
This dog was scared.

最佳答案

首先，实际上有一种不那么笨拙的方法。我们要做的就是改变print打印，对吧？

_print = print
def print(*args, **kw):
    args = (arg.replace('cat', 'dog') if isinstance(arg, str) else arg
            for arg in args)
    _print(*args, **kw)

import types

def print_function():
    print ("This cat was scared.")

def main():
    # A function object is a wrapper around a code object, with
    # a bit of extra stuff like default values and closure cells.
    # See inspect module docs for more details.
    co = print_function.__code__
    # A code object is a wrapper around a string of bytecode, with a
    # whole bunch of extra stuff, including a list of constants used
    # by that bytecode. Again see inspect module docs. Anyway, inside
    # the bytecode for string (which you can read by typing
    # dis.dis(string) in your REPL), there's going to be an
    # instruction like LOAD_CONST 1 to load the string literal onto
    # the stack to pass to the print function, and that works by just
    # reading co.co_consts[1]. So, that's what we want to change.
    consts = tuple(c.replace("cat", "dog") if isinstance(c, str) else c
                   for c in co.co_consts)
    # Unfortunately, code objects are immutable, so we have to create
    # a new one, copying over everything except for co_consts, which
    # we'll replace. And the initializer has a zillion parameters.
    # Try help(types.CodeType) at the REPL to see the whole list.
    co = types.CodeType(
        co.co_argcount, co.co_kwonlyargcount, co.co_nlocals,
        co.co_stacksize, co.co_flags, co.co_code,
        consts, co.co_names, co.co_varnames, co.co_filename,
        co.co_name, co.co_firstlineno, co.co_lnotab,
        co.co_freevars, co.co_cellvars)
    print_function.__code__ = co
    print_function()

main()

import ctypes
import internals # https://github.com/abarnert/superhackyinternals/blob/master/internals.py

def print_function():
    print ("This cat was scared.")

def main():
    for c in print_function.__code__.co_consts:
        if isinstance(c, str):
            idx = c.find('cat')
            if idx != -1:
                # Too much to explain here; just guess and learn to
                # love the segfaults...
                p = internals.PyUnicodeObject.from_address(id(c))
                assert p.compact and p.ascii
                addr = id(c) + internals.PyUnicodeObject.utf8_length.offset
                buf = (ctypes.c_int8 * 3).from_address(addr + idx)
                buf[:3] = b'dog'

    print_function()

main()

>>> n = 2
>>> pn = PyLongObject.from_address(id(n))
>>> pn.ob_digit[0]
2
>>> pn.ob_digit[0] = 1
>>> 2
1
>>> n * 3
3
>>> i = 10
>>> while i < 40:
...     i *= 2
...     print(i)
10
10
10

关于python - 是否有可能 "hack"Python 的打印功能？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/49271750/