nested-function

问题 I'm kind of new in C programming. To be honest this is my first program where I am using Function. But it's not working. Can anyone tell me what is supposed to be the problem? // Convert freezing and boiling point of water into Fahrenheit and Kelvin #include<stdio.h> void calculation(int); void freezingCalculation(void); void boilingCalculation(void); int main (){ int temperature, fahrenheit, kelvin; void freezingCalculation(void){ int temperature = 0; calculation(temperature); printf(

问题 I have a simple context that sets some value that it get from backend, pseudo code: export const FooContext = createContext(); export function Foo(props) { const [value, setValue] = useState(null); useEffect(() => { axios.get('/api/get-value').then((res) => { const data = res.data; setValue(data); }); }, []); return ( <FooContext.Provider value={[value]}> {props.children} </FooContext.Provider> ); } function App() { return ( <div className="App"> <Foo> <SomeView /> </Foo> </div> ); } function

问题 I have a simple context that sets some value that it get from backend, pseudo code: export const FooContext = createContext(); export function Foo(props) { const [value, setValue] = useState(null); useEffect(() => { axios.get('/api/get-value').then((res) => { const data = res.data; setValue(data); }); }, []); return ( <FooContext.Provider value={[value]}> {props.children} </FooContext.Provider> ); } function App() { return ( <div className="App"> <Foo> <SomeView /> </Foo> </div> ); } function

来源： https://stackoverflow.com/questions/13784998/js-nested-functions-gain-a-copy-of-the-outer-functions-parameters

来源： https://stackoverflow.com/questions/13784998/js-nested-functions-gain-a-copy-of-the-outer-functions-parameters

问题 This question already has answers here : Running unit tests on nested functions (6 answers) Closed 3 years ago . How would you unit test the nested function f1() in the following example? def f(): def f1(): return 1 return 2 Or should functions that need testing not be nested? 回答1: There is a similar question in this link. But short answer: you can't access an inner function from an outer element. For testing purposes, maybe an alternative would be to change the inner function for a private

问题 I have defined a function called initials inside a function called person , but I can't figure out how to call initials outside of person : main = --I attempted to print the output of the initials function here. (putStrLn ((person "firstName" "lastName") . initials)) --Not in scope: `initials' --If this function call worked correctly, the output would be "f.l.". person firstName lastName = firstName ++ ["."] ++ lastName where fullName = firstName ++ " " ++ lastName firstInitial = firstName !!

问题 I know this is the realm of closures and what not. But is it possible to continuously call nested anonymouse funtions? Say I have this: function testing(input) { var testing = 0; (function() { testing = testing + 1; })() return "testing"; } Can we have something like this testing()()()()()()() ? 回答1: You could use an inner function which makes the update and has a toString method to get a primitive value. function testing() { function fn() { string += ++counter; return fn; } var counter = 0,

问题 i am trying to call a function which is nested under other function consider a example: function _get_stats() { function _get_string() { $string = 'Nested Function Not Working'; return $string; } } public function index() { $data['title'] = $this->_get_stats() . _get_string(); $this->load->view('home', $data); } now when i run the page in web browser blank page is displayed. any suggestion or help would be a great help for me.. thanks in advance 回答1: The function is not really nested, but

问题 I am trying to create a parser using boost's spirit qi parser. It is parsing a string that contains three types of values. A constant, a variable, or a function. The functions can be nested inside of each other. The test string is f(a, b) = f(g(z, x), g(x, h(x)), c) , where a-e are constants, f-r are functions, and s-z are variables. I successfully created a rule that can correctly parse the expression. The trouble arose when I changed the function parsing the rule into a grammar. There were