Question

我最近一直在努力学习C ++并拿起了“C ++ Through Game Programming”这本书。我正在关于指针的章节，我已经有一个例子，我有一个问题。代码是这样的：

#include "stdafx.h"
#include <iostream>
using namespace std;

void badSwap(int x, int y);
void goodSwap(int* const pX, int* const pY);

int main()
{
    int myScore = 150;
    int yourScore = 1000;
    cout << "Original values\n";
    cout << "myScore: " << myScore << "\n";
    cout << "yourScore: " << yourScore << "\n\n";
    cout << "Calling badSwap()\n";
    badSwap(myScore, yourScore);
    cout << "myScore: " << myScore << "\n";
    cout << "yourScore: " << yourScore << "\n\n";
    cout << "Calling goodSwap()\n";
    goodSwap(&myScore, &yourScore);
    cout << "myScore: " << myScore << "\n";
    cout << "yourScore: " << yourScore << "\n";
    cin >> myScore;
    return 0;
}

void badSwap(int x, int y)
{
    int temp = x;
    x = y;
    y = temp;
}
void goodSwap(int* const pX, int* const pY)
{
    //store value pointed to by pX in temp
    int temp = *pX;
    //store value pointed to by pY in address pointed to by pX
    *pX = *pY;
    //store value originally pointed to by pX in address pointed to by pY
    *pY = temp;
}

在goodSwap（）函数中有一行：

*pX = *pY;

为什么要取消引用作业的双方？这不等于说“1000 = 150”吗？

Answer 1

为什么要取消引用作业的双方？不等于说＆＃34; 1000 = 150＆＃34;？

不，如下所示：

int x = 1000;
int y = 150;

x = y;

不等于说＆＃34; 1000 = 150＆＃34;。您正在分配对象，而不是它当前包含的值。

以下正好相同（因为表达式*px是引用对象x的左值，而表达式*py是左值引用对象y;它们实际上是别名，而不是某些奇怪的，断开连接的对象版本和数字值：

int   x = 1000;
int   y = 150;
int* px = &x;
int* py = &y;

*px = *py;

Answer 2

* px = * py表示我们从py的地址到px的地址分配值not address。

Answer 3

翻阅书中的章节或购买另一本书：现在不需要在C ++中使用普通指针。

还有一个std::swap函数，用于处理C ++ isch。

关于指针和解除引用

3 个答案: