side-effects

问题 Wikipedia says that: In computer science, an operation, function or expression is said to have a side effect if it modifies some state variable value(s) outside its local environment, that is to say has an observable effect besides returning a value (the main effect) to the invoker of the operation. But how can we access a variable outside its local environment, can anyone explain this situation, side effect, main effect and sequence point comprehensibly? 回答1: A function is (should be) a

来源： https://stackoverflow.com/questions/41829618/functional-programming-where-does-the-side-effect-actually-happen

来源： https://stackoverflow.com/questions/41829618/functional-programming-where-does-the-side-effect-actually-happen

问题 I have a celery retry task that I would like to test that it retries until successful. Using mock's side_effect, I can fail it for a set number of executions and then passing None , clear the side effect. However, the method the task is calling doesn't execute at that point, it just doesn't have an exception. Is there a way to clear the side effect, and still have the method being mocked execute as normal? I can test that it is called 'x' number of times (ie. repeat until successful) and then

问题 Webpack 4 has added a new feature: it now supports a sideEffects flag in the package.json of the modules it is bundling. From Webpack 4: released today Over the past 30 days we have worked closely with each of the frameworks to ensure that they are ready to support webpack 4 in their respective cli’s etc. Even popular library’s like lodash-es, RxJS are supporting the sideEffects flag, so by using their latest version you will see instant bundle size decreases out of the box. From Webpack docs

问题 I've got an idea for caching that I'm beginning to implement: Memoizing functions and storing the return along with a hash of the function signature in Velocity. Using PostSharp, I want to check the cache and return a rehydrated representation of the return value instead of calling the function again. I want to use attributes to control this behavior. Unfortunately, this could prove dangerous to other developers in my organization, if they fall in love with the performance gain and start

问题 C++03 Standard 1.9/6 defines observable behavior : The observable behavior of the abstract machine is its sequence of reads and writes to volatile data and calls to library I/O functions. and then and then 1.9/7 defines side effects : Accessing an object designated by a volatile lvalue (3.10), modifying an object, calling a library I/O function, or calling a function that does any of those operations are all side effects, which are changes in the state of the execution environment. Is a side

问题 C++03 Standard 1.9/6 defines observable behavior : The observable behavior of the abstract machine is its sequence of reads and writes to volatile data and calls to library I/O functions. and then and then 1.9/7 defines side effects : Accessing an object designated by a volatile lvalue (3.10), modifying an object, calling a library I/O function, or calling a function that does any of those operations are all side effects, which are changes in the state of the execution environment. Is a side

问题 I just got my copy of Expert F# 2.0 and came across this statement, which somewhat surprised me: For example, when necessary, you can use side effects on private data structures allocated at the start of an algorithm and then discard these data structures before returning a result; the overall result is then effectively a side-effect-free function. One example of separation from the F# library is the library's implementation of List.map, which uses mutation internally; the writes occur on an

问题 I have written a code and i have to verify it on the online compiler for this purpose. But that compiler is showing following error.Attaching the image Also my code is as follows: def matrixflip(m,d): n=m x=len(n[0][0:]) if(d=="h"): for i in range(len(n)): for j in range(x-1): temp=n[i][j] n[i][j]=n[i][j+1] n[i][j+1]=temp print(n) elif(d=="v"): temp=[] for i in range(len(n)-1): temp=n[i][0:] n[i][0:]=n[i+1][0:] n[i+1][0:]=temp print(n) else: print(m) myl=[[1,2],[3,4]] matrixflip(myl,"h")