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Tue Mar 6 01:02:47 CET 2007 · https://mail.python.org/pipermail/python-dev/2007-March/071515.html

"Dino Viehland" <dinov at exchange.microsoft.com> wrote in message 
news:7AD436E4270DD54A94238001769C22276A624C7482 at DF-GRTDANE-MSG.exchange.corp.microsoft.com...
def a():
                x = 4
                y = 2
                def b():
                                print y, locals()
                print locals()
                b()

a()

in CPython prints:

{'y': 2, 'x': 4, 'b': <function b at 0x020726F0>}
2 {'y': 2}

I'm wondering if it's intentional that these don't print dictionaries w/ 
the same contents or if it's more an accident of the implementation.   In 
other words would it be reasonable for IronPython to promote all of the 
locals of a into b's dictionary when both a and b call locals?

==========================================
This version

def a():
    x = 4
    y = 2
    def b():
        print y, locals()
    print locals()
    return b

a()()

has essentially the same output, as it should.  Do you really want the 
binding of 'x' and 'b' to survive the a's return?  I see no reason why a's 
call of locals() should affect this either way.  Which is to say, why the 
compilation of b should be affected by the code that follows it.

This version also has the same output

def a():
    x = 4
    def b():
        print y, locals()
    y = 2
    print locals()
    return b

a()()

whereas this omits y from a's output, but not b's:

def a():
    x = 4
    def b():
        print y, locals()
    print locals()
    y = 2
    return b

a()()

and would also if b were called instead of returned, as in your version. 
So it would not make too much sense for the two printouts to match.

Terry Jan Reedy

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[Python-Dev] locals(), closures, and IronPython...