pre增量VS阵列中增加后 [英] Pre increment vs Post increment in array

问题描述

我是新来的StackOverflow，这是我的第一篇文章。我学习编程，我已经从C语言开始。我读让我们C的书。我正想通过这个节目在这本书中。

I am new to Stackoverflow and this is my first post. I am learning programming and I have started from C language. I was reading Let us C book. And I was going through this program in that book.

main( ) 
{ 
 int a[5] = { 5, 1, 15, 20, 25 } ; 
 int i, j, k = 1, m ; 
 i = ++a[1] ; 
 j = a[1]++ ; 
 m = a[i++] ; 
 printf ( "\n%d %d %d", i, j, m ) ; 
}

我的理解是，它会打印我的2 ，Ĵ为1 和米15

但不知何故，它被打印成我是3 ，Ĵ为2 和米15 ？为什么会这样呢？

But somehow it is printing as i as 3, j as 2 and m as 15? Why is it so?

下面是我的理解 -

Below is my understanding-

b = x++;
In this example suppose the value of variable ‘x’ is 5 then value of variable ‘b’ will be 5 because old value of ‘x’ is used.

b = ++y;
In this example suppose the value of variable ‘y’ is 5 then value of variable ‘b’ will be 6 because the value of ‘y’ gets modified before using it in a expression.

这有什么错在我的理解？

Is there anything wrong in my understanding?

推荐答案

您击中了要害。你的理解是正确的。 pre和后增量前pressions之间的区别就像是它的声音。 pre-增量是指设置或评估的前pression之前变量递增。后增量意味着前pression设置或评价，然后将变量被改变。人们很容易把它当做一个两个步骤。

You hit the nail on the head. Your understanding is correct. The difference between pre and post increment expressions is just like it sounds. Pre-incrementation means the variable is incremented before the expression is set or evaluated. Post-incrementation means the expression is set or evaluated, and then the variable is altered. It's easy to think of it as a two step process.

b = x++;

真的是：

b = x;
x++;

和

b = ++x;

真的是：

x++;
b = x;

编辑：你提供的例子（这可能扔你了）最棘手的部分是，有一个数组的索引，并将其值之间的巨大差异。

The tricky part of the examples you provided (which probably threw you off) is that there's a huge difference between an array index, and its value.

i = ++a[1];

这意味着递增存储在[1]，然后将其设置为变量i的值。

That means increment the value stored at a[1], and then set it to the variable i.

m = a[i++];

这一种手段一套米的[I]，然后递增i的值。两者之间的区别是pretty大的区别，可以在第一会比较混乱。

This one means set m to the value of a[i], then increment i. The difference between the two is a pretty big distinction and can get confusing at first.

二编辑：code分解

{ 
 int a[5] = { 5, 1, 15, 20, 25 } ; 
 int i, j, k = 1, m ; 
 i = ++a[1] ; 
 j = a[1]++ ; 
 m = a[i++] ; 
 printf ( "\n%d %d %d", i, j, m ) ; 
}

首先：

i = ++a[1];

在这一点上，我们知道一个[1] = 1（记住数组是零索引）。但是，我们首先增加了。因此，我= 2。

At this point we know a[1] = 1 (remember arrays are zero indexed). But we increment it first. Therefore i = 2.

j = a[1]++;

记住我们递增一个[1]之前，因此它是目前2.我们设置J = 2，然后将其递增到3。因此J = 2和现在一[1] = 3

Remember we incremented a[1] before, so it is currently 2. We set j = 2, and THEN incremented it to 3. So j = 2 and now a[1] = 3.

m = a[i++];

我们知道I = 2，所以我们需要设置M = A [2]，然后递增我。在这个前pression结束时，M = 15，且i = 3

We know i = 2. So we need to set m = a[2], and then increment i. At the end of this expression, m = 15, and i = 3.

在总之，

i = 3, j = 2, m = 15.

这篇关于pre增量VS阵列中增加后的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！