一个简单循环中有多少个基本操作？

发布于 2024-10-18 08:20:29 字数 226 浏览 9 评论 0原文

我有一堆代码来查找原始操作。问题是，网络上并没有关于这个主题的详细资源。在这个循环中：

for i:=0 to n do
  print test
end

我们到底有多少步？在我的第一个猜测中，我会说 n+1 ，考虑到 n 表示循环次数，1 表示打印。后来我想，也许是我不够精确。不是有每次循环都给i加1的操作吗？就此而言，我们有 n+n+1=2n+1。这是正确的吗？

原文

I have a bunch of code to find the primitive operations for. The thing is that there aren't really many detailed resources out on the web on the subject. In this loop:

for i:=0 to n do
  print test
end

How many steps do we really have? In my first guess I would say n+1 considering n for the times looping and 1 for the print. Then I thought that maybe I am not precise enough. Isn't there an operation even to add 1 to i in every loop?
In that matter we have n+n+1=2n+1. Is that correct?

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活泼老夫 2024-10-25 08:20:29

通过将循环重新转换为 while ，可以将循环分解为“原始操作”：

int i = 0;
while (i < n)
{
    print test;
    i = i + 1;
}

或者，更明确地说：

loop:
    if (i < n) goto done
    print test
    i = i + 1
    goto loop
done:

您可以看到，对于每次迭代，都有一个比较、一个增量、和一个goto。这只是循环开销。您必须添加循环中完成的任何工作。如果 print 被视为“原始操作”，那么您将有：

n+1 次比较
n 次调用 print
n 次增量
n+1 goto指令（完成后从循环中分支出来的指令）

现在，如何将所有指令转换为机器代码高度依赖于编译器、运行时库、操作系统和目标硬件。也许还有其他事情。

The loop can be broken down into its "primitive operations" by re-casting it as a while:

int i = 0;
while (i < n)
{
    print test;
    i = i + 1;
}

Or, more explicitly:

loop:
    if (i < n) goto done
    print test
    i = i + 1
    goto loop
done:

You can then see that for each iteration, there is a comparison, an increment, and a goto. That's just the loop overhead. You'd have to add to that whatever work is done in the loop. If the print is considered a "primitive operation," then you have:

n+1 comparisons
n calls to print
n increments
n+1 goto instructions (one to branch out of the loop when done)

Now, how that all gets converted to machine code is highly dependent on the compiler, the runtime library, the operating system, and the target hardware. And perhaps other things.

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