local

问题 I'm trying to convert a local image to Base64 string . I am not using any HTML and simply need javascript which references the image's path within the code. For instance, converting: C:\Users\Work\Desktop\TestImage.jpg into /9j/4AAQSkZJRgABAQEASABIAAD/4QBKRXhpZgAASUkqAAgAAAADABoBBQABAAAAMgAAABsBBQABAAAAOgAAACgBAwABAAAAAgAAAAAAAAAAVOoqgJaYAABU6iqAlpgA/+IMWElDQ19QUk9GSUxFAAEBAAAMSExpbm8CEAAAbW50clJHQiBYWVogB84AAgAJAAYAMQAAYWNzcE1TRlQAAAAASUVDIH.....etc... There are many posts like this but they

问题 I have installed React using create-react-app . It installed fine, but I am trying to load an image in one of my components ( Header.js , file path: src/components/common/Header.js ) but it's not loading. Here is my code: import React from 'react'; export default () => { var logo1 = ( <img //src="https://s3.amazonaws.com/codecademy-content/courses/React/react_photo-goose.jpg" src={'/src/images/logo.png'} alt="Canvas Logo" /> ); return ( <div id="header-wrap"> <div className="container

问题 I am passing my local-variables by reference to two lambda. I call these lambdas outside of the function scope. Is this undefined ? std::pair<std::function<int()>, std::function<int()>> addSome() { int a = 0, b = 0; return std::make_pair([&a,&b] { ++a; ++b; return a+b; }, [&a, &b] { return a; }); } int main() { auto f = addSome(); std::cout << f.first() << " " << f.second(); return 0; } If it is not, however, changes in one lambda are not reflected in other lambda. Am i misunderstanding pass

问题 Most C++ programmers like me have made the following mistake at some point: class C { /*...*/ }; int main() { C c(); // declares a function c taking no arguments returning a C, // not, as intended by most, an object c of type C initialized // using the default constructor. c.foo(); // compiler complains here. //... } Now while the error is pretty obvious once you know it I was wondering if there is any sensible use for this kind of local function declaration except that you can do it --

问题 Merged with Can a local variable's memory be accessed outside its scope?. int * ref () { int tmp = 100; return &tmp; } int main () { int * a = ref(); cout << *a << endl; } I know the function ref () is allocated stack space. It will get destroyed as soon as the function exits. So the complier will give warning information. But my question is why the returning result is still correct. 来源： https://stackoverflow.com/questions/2862494/c-warning-address-of-local-variable

问题 Hi can anybody please explain me why is this code snippet giving me StackOverflowError I really appreciate if you can explain what is happening when instanceObj initializing and calling ObjectTest constructor and java.lang.Object constructor. It seems to me ObjectTest constructor loop over and over.But I don't know exact reason? So any suggestion... public class ObjectTest { public ObjectTest() { } ObjectTest instanceObj = new ObjectTest(); public static void main(String[] args) { ObjectTest

问题 I need some help with this: Example: void method1{ MyObject obj1=new MyObject(); obj1.method1(); } I want to mock obj1.method1() in my test but to be transparent so I don't want make and change of code. Is there any way to do this in Mockito? 回答1: If you really want to avoid touching this code, you can use Powermockito (PowerMock for Mockito). With this, amongst many other things, you can mock the construction of new objects in a very easy way. 回答2: The answer from @edutesoy points to the

问题 This question already has answers here : Closed 6 years ago . Possible Duplicate: Why are global and static variables initialized to their default values? See the code, #include <stdio.h> int a; int main(void) { int i; printf("%d %d\n", a, i); } Output 0 8683508 Here 'a' is initialized with '0', but 'i' is initialized with a 'junk value'. Why? 回答1: Because that's the way it is, according to the C Standard . The reason for that is efficiency: static variables are initialized at compile-time ,

问题 This question already has answers here : Closed 6 years ago . Possible Duplicate: Why are global and static variables initialized to their default values? See the code, #include <stdio.h> int a; int main(void) { int i; printf("%d %d\n", a, i); } Output 0 8683508 Here 'a' is initialized with '0', but 'i' is initialized with a 'junk value'. Why? 回答1: Because that's the way it is, according to the C Standard . The reason for that is efficiency: static variables are initialized at compile-time ,

问题 This question already has answers here : Closed 6 years ago . Possible Duplicate: Why are global and static variables initialized to their default values? See the code, #include <stdio.h> int a; int main(void) { int i; printf("%d %d\n", a, i); } Output 0 8683508 Here 'a' is initialized with '0', but 'i' is initialized with a 'junk value'. Why? 回答1: Because that's the way it is, according to the C Standard . The reason for that is efficiency: static variables are initialized at compile-time ,