后前润肤 [英] post-pre increament

问题描述

亲爱的Al，
早上好.
对于for循环中的后无声运算符和前无声运算符，我面临一些概念上的疑问.
请遵循以下代码:

Dear Al,
Very Good morning.
I am facing some conceptual doubts in post increament and pre increament operators in for loop.
Kindly follow the code below:

for(int i=0;i<5;i++)
{
cout<<i<<endl;
}

输出:

output:

0
1
2
3
4

再次，

Again,

for(int j=0;j<5;++j)
{
cout<<j<<endl;
}

输出将相同.

我的问题是代码的处理是如何进行的?
++j& i++都具有相同的输出，怎么办?

Output will be same.

My question is how the processing of the code takes place?
++j & i++ both are possessing the same output, how?

推荐答案

++i和i++的副作用是相同-i的值增加.但是表达式的值不同.

The side-effect of ++i and i++ is the same - the value of i is incremented. However the value of the expression is different.

int x = 2;
int j = ++x; // now x == 3 and j == 3
int q = x++; // now x == 4 and q == 3

因此，前增量是使用前的增量，后增量是使用后的增量.

So pre-increment is increment before use, and post-increment is increment after use.

了解前和后增量的方式工作中，我们需要一个不同于您编写的示例；看这个:

To understand how pre and post increment work, we need an example different from the one you wrote; look at this:

int i = 0;
int j = i++;
int k = ++i;
printf("i = %d\nj = %d\nk = %d\n", i, j, k);

这将产生如下输出:

This will produce an output like:

i = 2
j = 0
k = 2

语句int j = i++;的意思是:首先将i的值分配给j，然后将i递增(与int j = i; i = i + 1;相同)
语句int k = ++i;的意思是:首先递增i，然后将i的值赋给k(与i = i + 1; int k = i;相同)

The statement int j = i++; means: first assign to j the value of i and after increment i (it is the same than int j = i; i = i + 1;)
The statement int k = ++i; means: first increment i and after assign to k the value of i (it is the same than i = i + 1; int k = i;)

for行中的第二个分号-
将在第一个循环主循环之后处理:)

因此，您可以感觉到++ x&的区别. x ++(--x& x--)
例如以下几行:

The part after the second semicolon in a for line -
will be processed after the first loop primary :)

So you can feel the difference of ++x & x++ (--x & x--)
for example by the following lines:

{
  // one result
  int i = 5;
  do {
    // after-call decrement
    printf("%d\n", i--);
  } while(i);
  // another result
  printf("\n");
  i = 5;
  do {
    // pre-call decrement
    printf("%d\n", --i); 
  } while(i);
}

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