lambda

问题 So, I just upgraded to Xcode 4.4, and I noticed in the changelog: Apple LLVM compiler supports additional C++11 features, including lambdas Which is awesome! So I got around to coding, and I found a few things out: Lambdas are assignable to Objective-C blocks: void (^block)() = []() -> void { NSLog(@"Inside Lambda called as block!"); }; block(); std::function can hold an Objective-C block: std::function<void(void)> func = ^{ NSLog(@"Block inside std::function"); }; func(); We cant assign an

问题 I find myself doing this sort of thing quite often:- EventHandler eh = null; //can't assign lambda directly since it uses eh eh = (s, args) => { //small snippet of code here ((SomeType)s).SomeEvent -= eh; } variableOfSomeType.SomeEvent += eh; Basically I only want to attach an event handler to listen for one shot from the event, I no longer want to stay attached after that. Quite often that "snippert of code" is just one line. My mind is going a bit numb, I'm sure there must be something I

问题 I find myself doing this sort of thing quite often:- EventHandler eh = null; //can't assign lambda directly since it uses eh eh = (s, args) => { //small snippet of code here ((SomeType)s).SomeEvent -= eh; } variableOfSomeType.SomeEvent += eh; Basically I only want to attach an event handler to listen for one shot from the event, I no longer want to stay attached after that. Quite often that "snippert of code" is just one line. My mind is going a bit numb, I'm sure there must be something I

问题 I find myself doing this sort of thing quite often:- EventHandler eh = null; //can't assign lambda directly since it uses eh eh = (s, args) => { //small snippet of code here ((SomeType)s).SomeEvent -= eh; } variableOfSomeType.SomeEvent += eh; Basically I only want to attach an event handler to listen for one shot from the event, I no longer want to stay attached after that. Quite often that "snippert of code" is just one line. My mind is going a bit numb, I'm sure there must be something I

问题 So, I saw someone using the reduce method to calculate the Fibonacci sequence. Here is his idea: (1,0) , (1,1) , (2,1) , (3,2) , (5,3) corresponds to 1, 1, 2, 3, 5, 8, 13, 21 ....... and the code looks like this def fib_reduce(n): initial =(1,0) dummy = range(n) fib_n = reduce(lambda prev ,b : (prev[0] + prev[1], prev[0]), dummy, initial) return fib_n[0] I understand the (prev[0] + prev[1] , prev[0]) which is like a, b = b, b + a . However, I don't understand what this b stands for ? May

问题 I'm trying to accumulate the numbers in a vector using a multiplication lambda. What is my error? I get 1 as the result, instead of 24 (= 1 2 3*4). My approach is as follows: std::function<float(float a, int x)> func; std::vector<int> m{ 1, 2, 3, 4 }; // <-- Multiply: 1*2*3*4 = 24 float accumulation = 1.0f; func = [&accumulation, &m](float a, int i) { accumulation = a * *m.begin()++; return accumulation; }; accumulation = accumulate(m.cbegin(), m.cend(), accumulation, func); 回答1: The

问题 I need to copy properties of one object to another, both objects can be of different type, can have properties of same name. These property can also be of complex type. I was able to achieve the copy feature for simple TYPE properties, how ever i am unable to achieve this for complex types.. like see below sample snippet [TestClass] public class PropAssignTest { [TestMethod] public void Test() { Test1 t1 = new Test1() { Prop1 = "test", TestName = new Test3 { Name = "santosh" } } ; Test2 t2 =

问题 I want to join a list of tuples within a dataframe. I have tried several methods of doing this within the dataframe with join and with lambda import pandas as pd from nltk import word_tokenize, pos_tag, pos_tag_sents data = {'Categories': ['animal','plant','object'], 'Type': ['tree','dog','rock'], 'Comment': ['The NYC tree is very big', 'NY The cat from the UK is small', 'The rock was found in LA.']} def posTag(data): data = pd.DataFrame(data) comments = data['Comment'].tolist()

问题 we have peek function on stream which is intermediate function which accepts consumer. Then in my case why doesn't it replace "r" with "x". peek should ideally used for debugging purpose but I was just wondering why didn't it worked here. List<String> genre = new ArrayList<String>(Arrays.asList("rock", "pop", "jazz", "reggae")); System.out.println(genre.stream().peek(s-> s.replace("r","x")).peek(s->System.out.println(s)).filter(s -> s.indexOf("x") == 0).count()); 回答1: Because peek() accepts a

问题 I understand the syntax of the Java8 lambda expressions but why does the following code work without a specific type declaration of x? Why is "baz" being printed? public class LambdaExpressions { interface Foo { void bar(Object o); } static void doo(Foo f) { f.bar("baz"); } public static void main(String[] args) { doo( x -> {System.out.println(x);}); } } 回答1: Since the interface is a standard functional interface It's a functional interface because it contains only one abstract method. This