从涉及承诺的函数返回非承诺值 [英] return a non promise value from a function involving promises

问题描述

我有一个名为"test_sheet"的函数，该函数应该返回一个值.然后将该值传递给测试器函数，该函数将告诉我是否通过了测试.
在我的"test_sheet"内部，我有一些由promise处理的异步操作.现在，如何从我的test_sheet函数返回一个(非承诺)值.

I have a function called "test_sheet" that is supposed to return a value. that value will then be passed to a tester function which will tell me if I passed or failed the test.
inside my "test_sheet" I have a few async operations which are handled by promises. now, how can I return a (non-promise) value from my test_sheet function.

function test_sheet()
{
   //all my logic will go here

   new Promise(function(resolve, reject)
   {
      //simulating an async operation
      setTimeout(() => resolve(true), 1000);
   })
   .then(function(res){return res});
}

function tester()
{
   //not allowed to change this function in any way
   if(test_sheet() == true)
       console.log("pass!");
   else
       console.log("fail!");
}

tester();

还有更好的方法吗?

推荐答案

嗯，从技术上讲， tester()可能保持原样:

Well, technically it is possible, tester() may reamain intact:

var test_sheet=false;
function start_test()
{
   //all my logic will go here

   new Promise(function(resolve, reject)
   {
      //simulating an async operation
      setTimeout(() => resolve(true), 1000);
   })
   .then(res => {
      test_sheet=true;
      tester();
   });
}

function tester()
{
   //not allowed to change this function in any way
   test_sheet == true ? console.log("pass!") : console.log("fail!");
}

//tester();
start_test();

但是测试现在从 start_test()开始，并且 test_sheet 成为变量，其唯一目的是充当参数-不能添加到 testing()而无需对其进行修改.
这样，将无效的不良设计转换为不良的设计.

But the test starts with start_test() now, and test_sheet became a variable, with the sole purpose of acting as an argument - which could not be added to testing() without modifying it.
A nonworking bad design is transformed to working bad desing this way.

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