[ python-Bugs-1060135 ] incorrect "'yield' outside function" error

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Bugs item #1060135, was opened at 2004-11-04 06:25
Message generated for change (Comment added) made by rhettinger
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Category: Documentation
Group: None
>Status: Closed
>Resolution: Invalid
Priority: 1
Submitted By: Willem Broekema (willemb)
Assigned to: Nobody/Anonymous (nobody)
Summary: incorrect "'yield' outside function" error

Initial Comment:
In fact, the yield _is_ inside a function in the
following case:

>>> def f():
...   a = 3
...   exec("yield a")
...
>>> g = f()
Traceback (most recent call last):
  File "<stdin>", line 1, in ?
  File "<stdin>", line 3, in f
  File "<string>", line 1
SyntaxError: 'yield' outside function


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>Comment By: Raymond Hettinger (rhettinger)
Date: 2004-11-06 17:21

Message:
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Sorry.  No dice.
When you really understand what is happening,
the error message is dead-on.  You need to 
shift your mental model a bit.

The statements are executed without knowledge
of their surrounding context, only the dictionaries
can get shared.

When exec'd there is not enough information to
alter the message.

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Comment By: Willem Broekema (willemb)
Date: 2004-11-06 16:40

Message:
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It's the text of the error message that might be confusing;
something like "'yield' not allowed in 'exec'" would be better.

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Comment By: Raymond Hettinger (rhettinger)
Date: 2004-11-05 12:12

Message:
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Statements in an exec get interpreted and executed as a unit
and cannot affect surrounding control flow.  For example:
   for i in range(10):
       exec('continue')   # won't work

This is an intrinsic limitation and not a bug.  Perhaps the
docs could be improved though.

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