scope

问题 I'm trying to set the variables lat and lng to the coordinates returned by the geocoder. I think that lat and lng are accessible from inside the geocoder code where it says Geocoder.getFromLocation(inputWhere).then(...), but then outside of that, it doesn't seem to have set the variables to the geocoder's coordinates. Am I missing something about the variable scope? I've commented what the alerts are returning. I've tested that the geocoder works fine and that the "if" condition is being

问题 Elementor Pro (the WordPress page builder) integrates beautifully with Swiper, tying their GUI to the JS parameters and database content. However, for my project, I need to make some changes to the Swiper "CoverFlow" skin Init parameters (show more slides, change the 3D effect facing direction...). My hope is to to use the Destroy method of the Swiper API which looks like: mySwiper.destroy(deleteInstance, cleanStyles); Then I can initialize the Swiper again, with my own custom parameters. The

问题 The problem is C++ does not allow namespace declarations within classes.(I searched on internet and found this; if it's not true, please say it) So, what's the best way to cheat this problem? Context: My class have an enumeration within it. class GameCharacter { public: enum MovingState { Running, Walking, Stopped }; ... }; OBS: This example is not real, it's totally hypothetical. C++ defines that the enumeration names are inside the class scope, then to use these states I have to use the

问题 I thought this is how you make a variable inside a if/else block global (Case 3). connection.query(sql, function (err,rows){ //Handle Errors if(err){ //Log The Error & Die } //If Successful - Log Results For Now (Change To Export Results) else { //console.log(rows); foo = rows; } }); console.log(foo); // Out: undefined Why can't I access the variable outside of the function? SOLUTION Well, the problem here was one of understanding what asynchronicity really is . After chatting with @AmmarCSE

问题 //Main.c #include "Engine.h" int main() { Game app; //Menu Class Here app.run(); return 0; } //Engine.cpp #include "Engine.h" Game::Game() { DASH = 0, x = 0, y = 0; total_enemies = 1; p_timer = 0; e_timer = 0; } Game::~Game() { } void Game::run() { sf::RenderWindow window(sf::VideoMode(SCREEN_X, SCREEN_Y), "Shogun Master"); srand((unsigned int)time(NULL)); trying to replace the "creates player" here with void game::createPlayer that's located at the bottom of this file. //Creates Player

问题 I have the following simple code: template <typename T> struct base { std::vector<T> x; }; template <typename T> struct derived : base<T> { void print() { using base<T>::x; // error: base<T> is not a namespace std::cout << x << std::endl; } }; When I compile the code (using GCC-4.7.2) I get the error that you see in the comment above. I read here: http://gcc.gnu.org/onlinedocs/gcc-4.7.2/gcc/Name-lookup.html#Name-lookup that using base<T>::x has to be included in order to bring in the scope of

问题 This is an example from the book 'Matlab for Neuroscientists'. I don't understand the order in which, or why, g gets assigned a new value after each recursion. Nor do I understand why "factorial2" is included in the final line of code. here is a link to the text Basically, I am asking for someone to re-word the authors explanation (circled in red) of how the function works, as if they were explaining the concept and processes to a 5-year old. I'm brand new to programming. I thought I

问题 I've crossed an interesting problem. Suppose we have a class, and in its constructor we take a boolean as an argument. How can I define methods inside the class based on the instance's condition/boolean? For example: class X(): def __init__(self, x): self.x = x if self.x == true: # self is unreachable outside a method. def trueMethod(): print "The true method was defined." if self.x == false: # self is unreachable outside a method. def falseMethod(): print "The false method was defined." 回答1:

问题 I can not use the subset argument of any function with mapply . The following calls fail with the subset argument, but they work without: mapply(ftable, formula = list(wool ~ breaks, wool + tension ~ breaks), subset = list(breaks < 15, breaks < 20), MoreArgs = list(data = warpbreaks)) # Error in mapply(ftable, formula = list(wool ~ breaks, wool + tension ~ : # object 'breaks' not found mapply(xtabs, formula = list(~ wool, ~ wool + tension), subset = list(breaks < 15, breaks < 20), MoreArgs =

问题 I can not use the subset argument of any function with mapply . The following calls fail with the subset argument, but they work without: mapply(ftable, formula = list(wool ~ breaks, wool + tension ~ breaks), subset = list(breaks < 15, breaks < 20), MoreArgs = list(data = warpbreaks)) # Error in mapply(ftable, formula = list(wool ~ breaks, wool + tension ~ : # object 'breaks' not found mapply(xtabs, formula = list(~ wool, ~ wool + tension), subset = list(breaks < 15, breaks < 20), MoreArgs =