closures

问题 I'm trying to create a function on the fly that would return one constant value. In JavaScript and other modern imperative languages I would use closures: function id(a) { return function() {return a;}; } but Emacs lisp doesn't support those. I can create mix of identity function and partial function application but it's not supported either. So how do I do that? 回答1: Stupid idea: how about: (defun foo (x) `(lambda () ,x)) (funcall (foo 10)) ;; => 10 回答2: Found another solution with lexical

问题 I understand what closures are, but I am having some trouble grokking exactly what the term closure refers to. I have seen the term used in many websites, but rarely do they agree on the actual definition of it. Is it the variables that are kept on the stack frame? Is it the function that is being returned? Is it the scope of the outer function? Is it the scope of the inner (returned) function? Is it maybe the concept of keeping the variables on the stack-frame after returning the function?

问题 I understand what closures are, but I am having some trouble grokking exactly what the term closure refers to. I have seen the term used in many websites, but rarely do they agree on the actual definition of it. Is it the variables that are kept on the stack frame? Is it the function that is being returned? Is it the scope of the outer function? Is it the scope of the inner (returned) function? Is it maybe the concept of keeping the variables on the stack-frame after returning the function?

问题 This question already has answers here : JavaScript closure inside loops – simple practical example (44 answers) Closed 5 years ago . I'm having a problem with some JavaScript code. Script setTimeout(function() { for (var i = 0; i < 5; i++) { setTimeout(function() { console.log(i); }, i * 200); } }, 200); Outputs 5, 5, 5, 5, 5 instead of 1, 2, 3, 4, 5 I can kind of understand why this doesn't work, but I was wondering if someone could explain to me what's happening and why it's not working!

问题 This question already has answers here : JavaScript closure inside loops – simple practical example (44 answers) Closed 5 years ago . I'm having a problem with some JavaScript code. Script setTimeout(function() { for (var i = 0; i < 5; i++) { setTimeout(function() { console.log(i); }, i * 200); } }, 200); Outputs 5, 5, 5, 5, 5 instead of 1, 2, 3, 4, 5 I can kind of understand why this doesn't work, but I was wondering if someone could explain to me what's happening and why it's not working!

问题 I would like to declare a lifetime for a closure in Rust, but I can't find a way to add a lifetime declaration. use std::str::SplitWhitespace; pub struct ParserError { pub message: String, } fn missing_token(line_no: usize) -> ParserError { ParserError { message: format!("Missing token on line {}", line_no), } } fn process_string(line: &str, line_number: usize) -> Result<(), ParserError> { let mut tokens = line.split_whitespace(); match try!(tokens.next().ok_or(missing_token(line_number))) {

问题 Closures are poor man's objects and vice versa. I have seen this statement at many places on the web (including SO) but I don't quite understand what it means. Could someone please explain what it exactly means? If possible, please include examples in your answer. 回答1: Objects are poor man's closures. Consider Java. Java is an object-oriented programming language with no language level support for real lexical closures. As a work-around Java programmers use anonymous inner classes that can

问题 Closures are poor man's objects and vice versa. I have seen this statement at many places on the web (including SO) but I don't quite understand what it means. Could someone please explain what it exactly means? If possible, please include examples in your answer. 回答1: Objects are poor man's closures. Consider Java. Java is an object-oriented programming language with no language level support for real lexical closures. As a work-around Java programmers use anonymous inner classes that can

问题 I have a following simple code: def get(): return [lambda: i for i in [1, 2, 3]] for f in get(): print(f()) As expected from my python knowledge, output is 3 - entire list will contain last value of i . But how this works internally? AFAIK, python variables are simply reference to objects, so first closure must enclose object first i reference - and this object is definitely 1, not 3 O_O. How it happens that python closure encloses variable itself instead of object this variable reference?

问题 I have a following simple code: def get(): return [lambda: i for i in [1, 2, 3]] for f in get(): print(f()) As expected from my python knowledge, output is 3 - entire list will contain last value of i . But how this works internally? AFAIK, python variables are simply reference to objects, so first closure must enclose object first i reference - and this object is definitely 1, not 3 O_O. How it happens that python closure encloses variable itself instead of object this variable reference?