[英]How to break from inside a callback in a for-loop
If I bind
the function find
in javascript, I can do the following: 如果我bind
的功能, find
在JavaScript中,我能做到以下几点:
const _array = [4,10,6,5,20];
const loop = Array.prototype.find.bind(_array);
const r = loop(function (x) {
return x === 6;
});
console.log(`Final result => ${r}`); // here prints: Final result => 6
As you can see, in the binded loop
function I have a callback returned from find
. 如您所见,在绑定 loop
函数中,我从find
返回了一个回调 。 Everything works and it's ok ... 一切正常,没关系...
But, trying to simulate something like that I ended with this: 但是,尝试模拟类似的结果,我以此结束:
function loop(a,callback) {
for(i=0;i<a.length;i++)
callback(a[i]);
};
const r = loop([4,10,6,5,20], function (x) {
console.log("x value", x);
return x===6; // need to break the loop function and return to 'r' the 'x' value
});
console.log(`Final result => ${r}`); // here would print the value of x, that would be 6
and I get: 我得到:
x value 4
x value 10
x value 6
x value 5
x value 20
undefined
what means that the return x===6
inside the r
function is not working correctly, because the for-loop
continues to the end. 这意味着r
函数内部的return x===6
不能正常工作,因为for-loop
一直持续到最后。
So, my question: 所以,我的问题是:
How can I break the loop
function when x===6
and return the value of x
? 当x===6
时如何中断loop
函数并返回x
的值?
Check what value is returned by the callback, and then decide whether to continue or not: 检查回调返回的值,然后决定是否继续:
function loop(a, callback) { for (let i = 0; i < a.length; i++) { const found = callback(a[i]); if (found) { return a[i]; } } } const r = loop([4,10,6,5,20], function (x) { console.log("x value", x); return x===6; }); console.log(`Final result => ${r}`);
You can also write find
using recursion 您还find
使用递归编写find
const find = (f, [ x, ...xs ]) => x === undefined ? null : f (x) === true ? x : find (f, xs) console.log ( find ( x => x > 8 , [ 5, 7, 9, 3, 1 ] ) // 9 , find ( x => x < 4 , [ 5, 7, 9, 3, 1 ] ) // 3 )
Instead of destructuring assignment, an index parameter can be used 代替破坏分配,可以使用索引参数
const find = (f, xs = [], i = 0) =>
i >= xs.length
? null
: f (xs[i]) === true
? xs[i]
: find (f, xs, i + 1)
In both cases, iteration through the array stops as soon as f
returns true 在这两种情况下,只要f
返回true,就停止对数组的迭代
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