为什么我的箭头函数没有返回值? [英] Why doesn't my arrow function return a value?
问题描述
我有一个看起来像这样的箭头函数(简化):
I have an arrow function that looks like this (simplified):
const f = arg => { arg.toUpperCase(); };
但是当我调用它时,我得到 undefined
:
But when I call it, I get undefined
:
console.log(f("testing")); // undefined
为什么?
示例:
const f = arg => { arg.toUpperCase(); };
console.log(f("testing"));
(注意:这是一个干净的,规范的dupetarget,用于上面箭头函数的特定问题。)
(Note: This is meant to be a clean, canonical dupetarget for the specific issue with arrow functions above.)
推荐答案
当您使用箭头功能的函数体版本时(使用 {}
),没有隐含的返回
。你必须指定它。当您使用简洁正文(没有 {}
)时,函数会隐式返回正文表达式的结果。
When you use the function body version of an arrow function (with {}
), there is no implied return
. You have to specify it. When you use the concise body (no {}
), the result of the body expression is implicitly returned by the function.
所以你要写一个明确的返回
:
So you would write that either with an explicit return
:
const f = arg => { return arg.toUpperCase(); };
// Explicit return ^^^^^^
或简洁的机构:
const f = arg => arg.toUpperCase();
示例:
const f1 = arg => { return arg.toUpperCase(); };
console.log(f1("testing"));
const f2 = arg => arg.toUpperCase();
console.log(f2("testing"));
稍微切线,但说到 {}
:如果你想要简明扼要arrow的body表达式是一个对象初始值设定项,把它放在()
:
Slightly tangential, but speaking of {}
: If you want the concise arrow's body expression to be an object initializer, put it in ()
:
const f = arg => ({prop: arg.toUpperCase()});
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