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pythonsimpy

dynamically create simpy.resource

I am currently creating simpy.resource using the following command:

str1_counter = simpy.Resource(env, capacity=1)

But if I want to create it in a dynamic manner based on user input i.e. lets say user says the capacity(variable cap) should be 4, so in that case I tried the following:

cap = input("Enter the capacity of the store")
str1_counter = 'simpy.Resource(env, capacity=' + cap + ')'

But as you can see that instead of an instance of simpy.resource being created, the result is a string called 'simpy.Resource(env, capacity=4)' although I would have wanted simpy.resource(env,capacity=4).

How do I do this?

Solution

  • You could do it as such;

    cap = input("Enter the capacity of the store")
str1_counter = simpy.Resource(env, capacity=cap)

    But also be aware of checking the input and catching errors.. for example if someone enters five instead of 5, what should happen?