If you want to set a value, use the set method. get Methods are only to obtain variables, not to set the inner variables of a class (If they are defined the way you did). The correct usage is: Customer* CreateCustomer(const string& id, const string& name, const string& address) { Customer* temp = new Customer(); … Read more
Functions in Rust don’t capture variables from the surrounding environment, period. A “local” function in Rust is really just a global function that isn’t globally visible; it can’t do anything more than any other global function. Instead, Rust has closures which are distinct from functions in that they do capture variables from their environment. That … Read more
It’s just the way you think it would be, apply accepts args and kwargs and passes them directly to some_func. df.apply(some_func, var1=’DOG’, axis=1) Or, df.apply(some_func, args=(‘DOG’, ), axis=1) 0 foo-x-DOG 1 bar-y-DOG dtype: object If for any reason that won’t work for your use case, then you can always fallback to using a lambda: df.apply(lambda … Read more
Using a return inside of a loop will break it and exit the function even if the iteration is still not finished. For example: def num(): # Here there will be only one iteration # For number == 1 => 1 % 2 = 1 # So, break the loop and return the number for … Read more
The one thing you could test for is code object equality: >>> x = lambda x: x >>> y = lambda y: y >>> x.__code__.co_code ‘|\x00\x00S’ >>> x.__code__.co_code == y.__code__.co_code True Here the bytecode for both functions is the same. You’ll perhaps need to verify more aspects of the code objects (constants and closures spring … Read more
A first-class function is not a particular kind of function. All functions in Python are first-class functions. To say that functions are first-class in a certain programming language means that they can be passed around and manipulated similarly to how you would pass around and manipulate other kinds of objects (like integers or strings). You … Read more
No, Standard C does not support either. Why do you feel you need to convert your C++ code to C? That could get quite tricky – I’d have thought writing wrappers would be the way to go, if your C++ must be callable from C.
There are four main cases where you should use pass-by-reference over pass-by-value: If you are calling a function that needs to modify its arguments, use pass-by-reference or pass-by-pointer. Otherwise, you’ll get a copy of the argument. If you’re calling a function that needs to take a large object as a parameter, pass it by const … Read more
You want this: typedef void (K::*MemFuncType)() const; If you want to still base MemFuncType on FuncType, you need to change FuncType: typedef void FuncType() const; typedef FuncType K::* MemFuncType;