在for循环中计数 [英] Counting down in for-loops

问题描述

我相信（从一些研究阅读），在for循环倒计时实际上是更有效和更快的运行时间。我的完整的软件代码是C ++

我现在有这个：

  for（i = 0; i< domain; ++ i）{
  

my'i'is unsigned resgister int，
也是'domain'是unsigned int

用于遍历数组，例如

  array [i] = do stuff 
  

$ b $我可以想象这个答案是相当微不足道的，但是我不能得到这个答案。我的头圆。

更新：'做东西'不依赖于以前或以后的迭代。 for循环中的计算与i的迭代无关。 （我希望这是有道理的）。

更新：为了实现运行时加速与我的for循环，我倒计数，如果是这样删除未签名的部分，当我delcaring int，或者其他什么方法？

请帮忙解决方案

我猜你向后的循环看起来像这样：

  for（i = domain  -  1; i> = 0; --i）{
  

在这种情况下，因为 i 是无符号，它总是 大于或等于零。当你递减一个等于零的无符号变量时，它将绕回一个非常大的数字。解决办法是让 i 签名，或者像这样改变for循环中的条件：

  for（i = domain-1; i> = 0&& i< domain; --i）{
  域到 1 ，而不是从 domain  -  1  to  0 ： 
 
 
 <$ （i = domain; i> = 1;  -  i）{
 array [i-1] = ...; //注意你必须从循环里面减去1 
 
 $ $ $ 
 
I believe (from some research reading) that counting down in for-loops is actually more efficient and faster in runtime. My full software code is C++

I currently have this:
for (i=0; i<domain; ++i) {
my 'i' is unsigned resgister int,
also 'domain' is unsigned int

in the for-loop i is used for going through an array, e.g.
array[i] = do stuff
converting this to count down messes up the expected/correct output of my routine.

I can imagine the answer being quite trivial, but I can't get my head round it.

UPDATE: 'do stuff' does not depend on previous or later iteration. The calculations within the for-loop are independant for that iteration of i. (I hope that makes sense).

UPDATE: To achieve a runtime speedup with my for-loop, do I count down and if so remove the unsigned part when delcaring my int, or what other method?

Please help.
解决方案
I'm guessing your backward for loop looks like this:
for (i = domain - 1; i >= 0; --i) {
In that case, because i is unsigned, it will always be greater than or equal to zero. When you decrement an unsigned variable that is equal to zero, it will wrap around to a very large number. The solution is either to make i signed, or change the condition in the for loop like this:
for (i = domain - 1; i >= 0 && i < domain; --i) {
Or count from domain to 1 rather than from domain - 1 to 0:
for (i = domain; i >= 1; --i) {
    array[i - 1] = ...; // notice you have to subtract 1 from i inside the loop now
}


                        
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