[issue22091] __debug__ in compile(optimize=1)

[eryksun]

eryksun added the comment:

If __debug__ were referenced from the code object's co_consts instead of checking locals, globals and builtins (LOAD_NAME), then optimize=1 would work consistently for a given code object. 

Currently in 3.4.1:

    >>> dis.dis(compile("if __debug__: x = 1", "", "exec", optimize=0))
      1           0 LOAD_CONST               0 (1)
                  3 STORE_NAME               0 (x)
                  6 LOAD_CONST               1 (None)
                  9 RETURN_VALUE

    >>> dis.dis(compile("if __debug__: x = 1", "", "exec", optimize=1))
      1           0 LOAD_CONST               0 (None)
                  3 RETURN_VALUE

    >>> dis.dis(compile("x = __debug__", "", "exec", optimize=0))
      1           0 LOAD_NAME                0 (__debug__)
                  3 STORE_NAME               1 (x)
                  6 LOAD_CONST               0 (None)
                  9 RETURN_VALUE

    >>> dis.dis(compile("x = __debug__", "", "exec", optimize=1))
      1           0 LOAD_NAME                0 (__debug__)
                  3 STORE_NAME               1 (x)
                  6 LOAD_CONST               0 (None)
                  9 RETURN_VALUE

With the current design, an exec can override builtins.__debug__ in globals or locals:

    >>> exec("print(__debug__)", globals(), {"__debug__": "spam"})
    spam

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nosy: +eryksun

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Python tracker <report at bugs.python.org>
<http://bugs.python.org/issue22091>
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