No. You can try
def f():
var('whatever')
f()
print whatever
Nothing to do with the fact that 't' was globally available before.
By design the function var:
- creates a new symbolic variable (*not* a Python variable)
- makes it available in the global namespace as a Python variable
under the same name
Some more examples:
t = 5
-> creates a Python variable named t which contains 5
t = SR.var('x')
-> creates a Python variable named t which contains a symbolic
variable named x
var('t')
-> creates a Python variable named t which contains a symbolic
variable named t.
Indeed the latter should really be thought as
t = SR.var('t')
Vincent
On 21/12/15 14:35, Carl Eberhart wrote:
Ah. Thanks very much for that clarification.
Actually, my snippet illustrates the dilemma I was in.
t already has a value outside of f
executing f changes the value of t outside of f
that is what I would expect to happen if t were declared global in f, but I
thought t was local in f
I still love var, but now I know when to use SR.var instead
Carl
On Mon, Dec 21, 2015 at 9:49 AM, Jeroen Demeyer <jdeme...@cage.ugent.be>
wrote:
On 2015-12-21 16:38, Carl Eberhart wrote:
I admit I don't understand what is happening in the following snippit:
def f():
t=var('t')
t=5
a=2*t
return a
Solution: never use var() in a function. If you do need a symbolic
variable in a function (note that you don't in the snippet above), you can
use SR.var() instead of plain var(). That behaves like var(), except that
it does not change any global. Example:
sage: SR.var('y')
y
sage: y
NameError: name 'y' is not defined
You can use it with explicit assignment:
sage: y = SR.var('y')
sage: y
y
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