x++输出结果的有关问题，求大神帮忙解释下
x++输出结果的问题，求大神帮忙解释下
重点在void inPut()函数中，定义的int x=1，然后for循环中输出了x++，为何第一个输出还是“输入了1个数”，而不是2个？
求解释。
运行结果如下：
------解决思路----------------------
看一下++i与i++就明白了
------解决思路----------------------
Postfix Increment and Decrement Operators: ++, --
postfix-expression :
postfix-expression ++
postfix-expression --
Operands of the postfix increment and decrement operators are scalar types that are modifiable l-values.
The result of the postfix increment or decrement operation is the value of the postfix-expression before the increment or decrement operator is applied. The type of the result is the same as that of the postfix-expression but is no longer an l-value. After the result is obtained, the value of the operand is incremented (or decremented).
For more information, see Prefix Increment and Decrement Operators.
Example
In the following example, the for loop is iterated 10 times. After each iteration, the nCount variable increased by +1 using the postfix increment operator.
// Example of the postfix increment operator
for (nCount = 0; nCount < 10; nCount++) {
nTotal += nArray[Count];
}
------解决思路----------------------
cout << "输入了" << x++<< "个数\n";
如果写的是i，那么第一次会输出0，你写的x，那第一次会输出你的初始值1，i++属于先使用再加加的意思，就是先输出，再自增
------解决思路----------------------
说通俗点，x = 1时：a = x++中， a 的值仍是1，而x的值才是2，而++x则都是2
// 7.11.cpp : 定义控制台应用程序的入口点。
//冒泡排序
#include "stdafx.h"
#include<iostream>
#include<stdlib.h>
using namespace std;
#define N 5
void inPut(int *a) //输入要比较的数
{
int x = 1;
for (int i = 0; i < N; i++)
{
cin >> a[i];
cout << "输入了" << x++<< "个数\n";
}
}
void print(int *a) //输出数组的数
{
for (int i = 0; i < N; i++)
{
cout << a[i] << " ";
cout << endl;
}
}
void bubble(int *a) //冒泡排序
{
int t = 0;
for (int i = 0; i < N - 1; i++)
{
for (int j = 0; j < N - i - 1; j++)
{
if (a[j]>a[j + 1])
{
t = a[j];
a[j] = a[j + 1];
a[j + 1] = t;
}
}
}
}
int _tmain(int argc, _TCHAR* argv[])
{
int a[N] = { 0 }; //定义一个整型数组a
cout << "请输入要排序的数：\n";
inPut(a); //输入
cout << "\n输入的"<<N<<"个数为：\n";
print(a); //输入排序前
bubble(a);
cout << "排序后：\n";
print(a);
system("pause");
return 0;
}
重点在void inPut()函数中，定义的int x=1，然后for循环中输出了x++，为何第一个输出还是“输入了1个数”，而不是2个？
求解释。
运行结果如下：
------解决思路----------------------
看一下++i与i++就明白了
------解决思路----------------------
Postfix Increment and Decrement Operators: ++, --
postfix-expression :
postfix-expression ++
postfix-expression --
Operands of the postfix increment and decrement operators are scalar types that are modifiable l-values.
The result of the postfix increment or decrement operation is the value of the postfix-expression before the increment or decrement operator is applied. The type of the result is the same as that of the postfix-expression but is no longer an l-value. After the result is obtained, the value of the operand is incremented (or decremented).
For more information, see Prefix Increment and Decrement Operators.
Example
In the following example, the for loop is iterated 10 times. After each iteration, the nCount variable increased by +1 using the postfix increment operator.
// Example of the postfix increment operator
for (nCount = 0; nCount < 10; nCount++) {
nTotal += nArray[Count];
}
------解决思路----------------------
cout << "输入了" << x++<< "个数\n";
如果写的是i，那么第一次会输出0，你写的x，那第一次会输出你的初始值1，i++属于先使用再加加的意思，就是先输出，再自增
------解决思路----------------------
说通俗点，x = 1时：a = x++中， a 的值仍是1，而x的值才是2，而++x则都是2