Javascript 如何反转 FOR 循环中的顺序
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How to reverse the order in a FOR loop
提问by Raspo
I've a simple FOR statement like this:
我有一个简单的 FOR 语句，如下所示：
var num = 10,
reverse = false;
for(i=0;i<num;i++){
console.log(i);
}
when reverseis false I want it to return something like [0,1,2,3,4,5,6,7,8,9]
当reverse为 false 我希望它返回类似 [0,1,2,3,4,5,6,7,8,9]
but, when reverseis true, it should return [9,8,7,6,5,4,3,2,1,0]
但是，当reverse为真时，它应该返回 [9,8,7,6,5,4,3,2,1,0]
Which is the most efficient way to get this result, without checking every time if reverseis true or false inside the loop?
哪个是获得此结果的最有效方法，无需每次都检查循环内的reverse是真还是假？
I don't want to do this:
我不想这样做：
var num = 10,
reverse = false;
for(i=0;i<num;i++){
if(reverse) console.log(num-i)
else console.log(i)
}
I would like to check reverseonly one time outside the loop.
我只想在循环外检查一次反向。
回答by user113716
var num = 10,
reverse = false;
if(!reverse) for( var i=0;i<num;i++) log(i);
else while(num-- ) log(num);
// to avoid duplication if the code gets long
function log( num ) { console.log( num ); }
EDIT:
编辑：
As noted in the comments below, if iis not declared elsewhere and you do not intend for it to be global, then declare it with the other variables you declared.
正如下面的评论中所指出的，如果i没有在其他地方声明并且您不打算将其设为全局，则使用您声明的其他变量声明它。
And if you don't want to modify the value of num, then assign it to ifirst.
如果您不想修改 的值num，则将其分配给i第一个。
var num = 10,
reverse = false,
i;
if(!reverse) for(var i=0;i<num;i++) log(i); // Count up
else {var i=num; while(i--) log(i);} // Count down
function log( num ) { console.log( num ); }
回答by Ivan Nevostruev
Try use 2 loops:
尝试使用 2 个循环：
if (reverse) {
for(i=num-1;i>=0;i--){
console.log(i)
}
}
else {
for(i=0;i<num;i++){
console.log(i)
}
}
回答by Roy Sharon
var num = 10,
reverse = false;
for (var i = 0, n = reverse?num-1:0, d = reverse?-1:1; i < num; i++, n+=d) {
console.log(n);
}
This is equivalent to the following, which is more readable, but less compact:
这等效于以下内容，它更具可读性，但不那么紧凑：
var num = 10,
reverse = false;
var start = reverse ? num-1 : 0,
end = reverse ? -1 : num,
step = reverse ? -1 : 1;
for (var i = start; i != end; i += step) {
console.log(i);
}
Edit:
Actually, these two solutions are not identical, because the first one has an additional increment operation. Still, it is negligible from performance point of view. If you really want to get a compact solution that has the best performance, you can do the following (not for the faint of heart):
编辑：
实际上，这两个解决方案并不相同，因为第一个解决方案有一个额外的增量操作。尽管如此，从性能的角度来看，它可以忽略不计。如果您真的想获得具有最佳性能的紧凑解决方案，您可以执行以下操作（不适合胆小的人）：
var num = 10,
reverse = false;
for (var r=reverse, i=r?num-1:0, n=r?-1:num, d=r?-1:1; i!=n; i+=d) {
console.log(i);
}
This has the advantage of having a single control structure, a single test in each loop, and a single iterator addition. It is not as fast as having an iterator increment/decrement, but only marginally so.
这具有单个控制结构、每个循环中的单个测试和单个迭代器添加的优点。它没有迭代器递增/递减那么快，但只是略微如此。
回答by Jerome
var start; var end; var inc;
if (reverse) {
start = num-1; end = 0; inc = -1;
}
else {
start = 0; end = num-1; inc = 1;
}
for(i=start;i!=end;i+=inc){
console.log(i)
}
回答by Jerome
I just came across the need for this the other day. Here's how I did it:
前几天我刚遇到这个需求。这是我如何做到的：
var num = 10,
i = 0,
direction = 1,
reverse = false;
if(reverse)
i = num + (direction = num = -1);
for(; i !== num; i += direction) {
console.log(i);
}
No need for separate loops, and no need to do math to calculate the proper iin the loop.
不需要单独的循环，也不需要做数学计算来计算i循环中的正确值。
So if reverseis true...
所以如果reverse是true...
i(which represents our first item) becomesnum - 1, so we're now starting on what would have been the last itemnum(which represents out of bounds) becomes-1, so we're now stopping on what would have been the first itemdirectionis-1, which means it will decrement when we doi += direction
i（代表我们的第一个项目）变成num - 1，所以我们现在从最后一个项目开始num（代表越界）变成-1，所以我们现在停在第一个项目上directionis-1，这意味着当我们这样做时它会递减i += direction
So by swapping our starting point with our ending point and changing the alteration of ifrom 1to -1, we'll be going up or down based on those modifications.
因此，通过将起点与终点交换并更改ifrom1到的更改-1，我们将根据这些修改上升或下降。
回答by Mike Dunlavey
Here's how I've always done reverse loops:
这是我一直以来做反向循环的方式：
for (i = num; --i >= 0; ) ...
回答by Bernard
I think this meets your requirements:
我认为这符合您的要求：
var num = 10;
var reverse = false;
var diff = 0;
if (reverse) {
diff = num - 1;
}
for (i = 0; i < num; i++) {
console.log(Math.abs(diff - i));
}
回答by XstreamINsanity
Roy's is similar to mine, but here's what I was thinking. I'll give you what I wrote in C# and how I think it translates to Javascript.
罗伊的和我的很相似，但这是我的想法。我会给你我用 C# 写的东西，以及我认为它是如何转换成 Javascript 的。
C#
C＃
int num = 10;
bool reverse = true;
for (int i = reverse ? num : 0; (reverse ? 0 : i) < (reverse ? i : num); i += reverse ? -1 : 1)
{
Console.Write((reverse ? i - 1 : i).ToString());
}
Console.ReadKey();
Javascript
Javascript
var num = 10,
reverse = true;
for (int i = reverse ? num : 0; (reverse ? 0 : i) < (reverse ? i : num); i += reverse ? -1 : 1)
{
console.log(reverse ? i - 1 : i);
}
And here's another way
Javascript
这是另一种方式
Javascript
var num = 10,
reverse = false;
for (int i = 0; i < num; i++)
{
console.log((reverse ? abs(-num + (i + 1)) : i));
}
回答by abele
It seems to work:
它似乎有效：
var num = 10;
var z = 1;
var k = -10;
if (reverse ){
k = -1;
z = -1;
}
for(int i = 0; i < 10; i++){
console.log(num+i*z+k);
}
回答by dash-tom-bang
Surely in a language like Javascript there must be a way to define a local function and use that in the loop?
当然，在像 Javascript 这样的语言中，必须有一种方法来定义本地函数并在循环中使用它吗？
function SubtractFrom(val, subtractor) {
return val - subtractor;
}
function PassThrough(val) {
return val;
}
var num = 10;
var processor = reverse ? SubtractFrom(num-1) : PassThrough;
for (i = 0; i < num; i++) {
console.log(processor(i));
}
Not knowing Javascript though, I don't know what actual form the function definitions would take.
虽然不知道 Javascript，但我不知道函数定义会采用什么实际形式。
