Python runs functions that are elements of a declared dictionary

Question

As soon as i define this dictionary:

case_dict={
    "run" : runn(),
    "name" : namee(),
    "help" : helpp(),
    "quit": quitt(),
}

python runs all these functions successively. How do i avoid these functions from running in the console when i just want to define the dict ?

PS: all these functions are defined earlier in the code.

When do you plan to execute them ? What exactly do you expact to happen? — adir abargil, Dec 17 '20 at 17:56
`runn` is the name of a function object and `runn()` is a call to it. — martineau, Dec 17 '20 at 17:58
I plan to execute them when i call them from the dict later on in the program — bendo97, Dec 17 '20 at 18:05

SyntaxError · Answer 1 · 2020-12-17T18:09:02.787

3

You assigned the value of each key to the return value of the function, not the function itself. Through the brackets at the end the functions get executed the moment you create the dict.

to avoid this, write runn instead of runn()

so you would end up with

case_dict={
    "run" : runn,
    "name" : namee,
    "help" : helpp,
    "quit": quitt,
}

Now, if you would like to call one of these functions you would do it like so:

case_dict["run"](parameter_1, parameter_2 ... )

edited Dec 17 '20 at 18:09

answered Dec 17 '20 at 17:56

SyntaxError

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yes, but when i call an element of the dict like: case_dict["run"] it doesn't run the function, it returns the runn function object. I want each function defined as an element of the dict to run when i call it from the dict. – bendo97 Dec 17 '20 at 18:03
yes, in order to call the function you always have to use brackets. so you would have to write `case_dict["run"]()` – SyntaxError Dec 17 '20 at 18:06
Ohh ok, now i get it. Thanks a lot ! – bendo97 Dec 17 '20 at 18:07
@bendo97 that isn't what "calling" means – juanpa.arrivillaga Dec 17 '20 at 18:24

Python runs functions that are elements of a declared dictionary

1 Answers1