问题
When I use Manipulate I can do:
Manipulate[x, {u, 1, 10}]
In reality my controls are many and complicated, so I would prefer to take their definition out of the Manipulate expression, like that:
control = {u, 1, 10}
Manipulate[x, control]
But that does result in a an error:
Manipulate argument control does not have the correct form for a \
variable specification.
Why doesn't it work that way?
回答1:
Manipulate has the HoldAll attribute. You can force control to evaluate and everything works ok
control = {u, 1, 10};
Manipulate[x[u], Evaluate[control]]
The problem with this is that the variable u is not properly localised, so if you have already set, e.g., u=1 somewhere, then the Manipulate will return an error.
It might be better if you use appropriate scoping constructs such as With or DynamicModule depending on exactly what you're trying to do.
This is maybe overkill, but it ensures that u is local and moves control outside of the manipulate:
DynamicModule[{u}, With[{control = {u, 1, 10}}, Manipulate[x[u], control]]]
回答2:
This
con = {u, 1, 10};
Manipulate[
u,
Evaluate@con
]
does work. I suppose it doesn't work without the Evaluate because
Attributes[Manipulate]
shows that Manipulate has the attribute HoldAll (but I may be wrong). To see the effect of this attribute, try this:
SetAttributes[f, HoldAll]
f[con]
f[Evaluate@con]
g[con]
(*
f[con]
f[{u, 1, 10}]
g[{u, 1, 10}]
*)
Thus, it appears that due to the HoldAll atribute, Manipulate simply does not see "inside" con unless you explicitly evaluate it.
来源：https://stackoverflow.com/questions/7138827/define-control-as-variable-in-mathematica