Yes bum is just what youâre calling the input you give to the function. When you invoke the function you are most certainly passing some value into the function. The function uses your declaration to see what you want to do with the input and then uses return as a way to output whatever you did to the data. You can call it whatever you want which is why whether it is a num or a bum it works.
Running timesFive(3) is the same as running the inner part of your function with bum = 3. Since the only line of code is: return bum * 5; which is: return 3 * 5; it will return 15.
No itâs because num is a parameter of the function. Itâs your way of passing input to the function to do something with it. In this case youâre taking the input and and multiplying it by 5. SInce the input you pass in is 3 (indicated when you invoked the function with the line âtimesFive(3);â), youâre really saying 3 * 5 which is 15.