Stefan Ram wrote:
When one defines a function, sometimes its name is only
half-existent.
One can implicitly evaluate the name of the function:
main.py
def g():
def f():
print( f )
f()
g()
output
<function g.<locals>.f at ...
, but one gets an error when one tries to evaluate it explicitly:
main.py
def g():
def f():
print( eval( 'f' ))
f()
g()
error output
NameError: name 'f' is not defined
I'm guessing that's because the name "f", referencing that function, is
in the local scope of the "g" function.
In the first version, the interpreter searches the scope of the code
which calls "print(f)". It doesn't find anything named "f" there, so
searches the next scope out, that of the function identified by "g".
That contains "f", so it's found and can be printed.
In the second case, eval() only gets the globals and immediate locals,
in this case the locals of "f". The name "f" doesn't exist in the local
scope of "f", and it doesn't exist in the global scope. So the code
executed by eval() can't see it.
The following does work:
def g():
def f():
print(eval('g'))
f()
g()
...because in this case "g" is defined in the global scope, so the code
in the eval call can see it.
The following also works:
def g():
def f():
pass
print(eval('f'))
g()
...because in this case, eval() is called from within the scope of "g",
so it can see the function "f" defined in that scope.
--
Mark.
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