name-binding

JAX-RS writer interceptor works for every response even with NameBinding

由 假如想象 提交于 2020-03-03 13:00:14
问题 I need to intercept a response from the server that was generated for a specific API call and change it. I used the JAX-RS 'WriterInterceptor' for this and I have created a NameBinding as well. But the server is intercepting every response out from the server. What am I doing wrong here? Shown below is the code I have tried. Why is the name binding does not work? (I have verified that when calling other API resources the particular method that I have use name binding is not called.) Name

Python; name bindings are not object references?

由 大城市里の小女人 提交于 2019-12-24 00:47:22
问题 I am trying to understand what exactly a Python name binding is, and when this binding is interpreted. In c, include <stdio.h> int main() { int X = 42; int* Y[1]; Y[0] = &X; X = 666; printf("%d", *Y[0]); return 0; } prints 666. I was expecting the block of Python code: X = 42 L = [] L.append(X) #3 X = 666 print(L) #5 to do the same, but it does not. What exactly happens between the lines labeled 3 and 5? Does #3 make another reference to the object known as "42", like X, lets call it X', and

JAX RS, my filter is not working

由 血红的双手。 提交于 2019-12-12 13:29:52
问题 Im work on : Best practice for REST token-based authentication with JAX-RS and Jersey But my filter not triggered, my call pass directly to the endpoint... My secure interface: @Qualifier @Retention(RUNTIME) @Target({METHOD, FIELD, PARAMETER, TYPE}) public @interface Secure { } My filter: @Secure @Provider @Priority(Priorities.AUTHENTICATION) public class AuthenticationFilter implements ContainerRequestFilter { @Override public void filter(ContainerRequestContext requestContext) throws

Python nested scopes with dynamic features

由 给你一囗甜甜゛ 提交于 2019-12-03 17:22:43
问题 Need help with understanding the following sentence from PEP 227 and the Python Language Reference If a variable is referenced in an enclosed scope, it is an error to delete the name. The compiler will raise a SyntaxError for 'del name'. Lack of examples caused I couldn't reproduce an error at compile time, so an explanation with examples is highly desirable. 回答1: The following raises the execption: def foo(): spam = 'eggs' def bar(): print spam del spam because the spam variable is being

point of instantiation and name binding

由 ∥☆過路亽．° 提交于 2019-11-30 22:52:59
I am confused about the point of instantiation with the following example: #include <iostream> void f(int){std::cout<<"int"<<std::endl;}//3 template <typename T> void g(T t) { f(t);//4 } void f(double){std::cout<<"double"<<std::endl;} int main() { g<int>(1);//1.point of instantiation for g<int> g<double>(1.1);//2.point of instantiation for g<double>, so f(double) is visible from here? return 0; } I though f is a dependent name and 1. is the point of instantiation for g< int > and 2. is the point of instantiation for g< double >, so f(double) is visible for g(1.1), however the output is int int

point of instantiation and name binding

由 血红的双手。 提交于 2019-11-30 17:25:08
问题 I am confused about the point of instantiation with the following example: #include <iostream> void f(int){std::cout<<"int"<<std::endl;}//3 template <typename T> void g(T t) { f(t);//4 } void f(double){std::cout<<"double"<<std::endl;} int main() { g<int>(1);//1.point of instantiation for g<int> g<double>(1.1);//2.point of instantiation for g<double>, so f(double) is visible from here? return 0; } I though f is a dependent name and 1. is the point of instantiation for g< int > and 2. is the

What happens when JavaScript variable name and function name is the same?

由 南笙酒味 提交于 2019-11-27 19:08:53
I have the following code, where I declare a function and after it, a variable with the same name as the function: function a(x) { return x * 2; } var a; alert(a); I expected this to alert undefined , but if I run it, the alert will display the following: function a(x) { return x * 2 } If I assign a value to the variable (like var a = 4 ), the alert will display that value ( 4 ), but without this change a will be recognized as a function. Why is this happening? Functions are a type of object which are a type of value . Values can be stored in variables (and properties, and passed as arguments

What happens when JavaScript variable name and function name is the same?

由 折月煮酒 提交于 2019-11-26 15:55:30
问题 I have the following code, where I declare a function and after it, a variable with the same name as the function: function a(x) { return x * 2; } var a; alert(a); I expected this to alert undefined , but if I run it, the alert will display the following: function a(x) { return x * 2 } If I assign a value to the variable (like var a = 4 ), the alert will display that value ( 4 ), but without this change a will be recognized as a function. Why is this happening? 回答1: Functions are a type of
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