当我有一个条件“if （bool = true）”时，else 分支永远不会被命中。为什么？-解网

问：

挑战如下：

创建一个接受字符串的函数 makePlans。此字符串应为名称。函数 makePlans 应调用函数 callFriend 并返回结果。callFriend 接受布尔值和字符串。将 friendsAvailable 变量和 name 传入 callFriend。

创建一个接受布尔值和字符串的函数 callFriend。如果布尔值为 true，则 callFriend 应返回字符串“本周末使用 NAME 制定的计划”。否则，它应该返回“这个周末每个人都很忙”>

这是我写的：

let friendsAvailable = true;

function makePlans(name) {
  return callFriend(friendsAvailable, name);
}

function callFriend(bool, name) {
  if (bool = true) {
    return 'Plans made with ' + (name) + ' this weekend'
  } else {
    'Everyone is busy this weekend'
  }

}

console.log(makePlans("Mary")) // should return: "Plans made with Mary this weekend'
friendsAvailable = false;
console.log(makePlans("James")) //should return: "Everyone is busy this weekend."

JavaScript 赋值运算符 equality-operator

let friendsAvailable = true;

function makePlans(name)
  {
  return callFriend(friendsAvailable, name);
  }

function callFriend(bool, name)
  {
  if (bool)  // or  if (bool===true), but testing if true is true is a little bit redundant
    {
    return 'Plans made with ' + (name) + ' this weekend'
    } 
  else 
    {
    return 'Everyone is busy this weekend'
    }
  }

console.log(makePlans("Mary")) // should return: "Plans made with Mary this weekend'
friendsAvailable = false;
console.log(makePlans("James")) //should return: "Everyone is busy this weekend."

但是，如果你想给你的老师留下深刻印象，请这样做：

const
  callFriend = 
    (bool, name) =>
      bool 
       ? `Plans made with ${name} this weekend` 
       : 'Everyone is busy this weekend' 

const makePlans = name => callFriend(friendsAvailable, name);


let friendsAvailable = true

console.log(makePlans('Mary')) 

friendsAvailable = false

console.log(makePlans('James'))

一些帮助程序：
箭头函数表达式
 条件（三元）运算符

当我有一个条件“if （bool = true）”时，else 分支永远不会被命中。为什么？

When I have a condition `if (bool = true)`, the else branch is never hit. Why?

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