根据答案https://stackoverflow.com/a/11842442/5835947，如果你这样编码，函数参数Bubble * targetBubble将被复制到函数内部。

bool clickOnBubble(sf::Vector2i & mousePos, std::vector<Bubble *> bubbles, Bubble * targetBubble) {
    targetBubble = bubbles[i];
}

然而，我做了一个测试，发现作为函数参数的指针将与外部的指针相同，直到我改变了它的值：

// c++ test ConsoleApplication2.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include "c++ test ConsoleApplication2.h"
using namespace std;
#include<iostream>



int main()
{
    int a= 1;
    int* pointerOfA = &a;

    cout << "address of pointer is" << pointerOfA << endl;
    cout << *pointerOfA << endl;
    func(pointerOfA);
    cout << *pointerOfA << endl;



}

void func(int *pointer)
{
    cout << "address of pointer is " << pointer  <<" it's the same as the pointer outside!"<<endl;

    int b = 2;
    pointer = &b;
    cout << "address of pointer is" << pointer <<" it's changed!"<< endl;

    cout << *pointer<<endl;

}

输出如下：

address of pointer is0093FEB4
1
address of pointer is 0093FEB4 it's the same as the pointer outside!
address of pointer is0093FDC4 it's changed!
2
1

所以，事实是，指针作为函数参数是不会被复制的，除非它被改变了，对吗？，或者我错过了什么？