Pass-by-reference in C++ and java

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本文通过示例详细对比了C++中引用传递与指针传递的区别，并解释了如何选择使用引用还是指针。此外，还探讨了Java中的参数传递机制，澄清了Java实际上是按值传递对象引用这一常见误解。

- pass-by-reference and pass-by-pointer in C++

Refer to this
Reference is An alias (an alternate name) for an object. In following sample(swap(int& x, int& y), x is a — not a pointer to a, nor a copy of a, but a itself. Anything you do to a gets done to a, and vice versa.

Use references when you can, and pointers when you have to. References are usually preferred over pointers whenever you don’t need “reseating”. This usually means that references are most useful in a class’s public interface. References typically appear on the skin of an object, and pointers on the inside.

sample code:

#include <iostream>
using namespace std;
void swap(int* x, int* y)
{
    int z = *x;
    *x=*y;
    *y=z;
}
void swap(int& x, int& y)
{
    int z = x;
    x=y;
    y=z;
}

int main()
{
    int a = 45;
    int b = 35;
    cout<<"Before Swap\n";
    cout<<"a="<<a<<" b="<<b<<"\n";

    swap(&a,&b);
    cout<<"After Swap with pass by pointer\n";
    cout<<"a="<<a<<" b="<<b<<"\n";

    swap(a,b);
    cout<<"After Swap with pass by reference\n";
    cout<<"a="<<a<<" b="<<b<<"\n";
}

Output:

Before Swap
a=45 b=35
After Swap with pass by pointer
a=35 b=45

After Swap with pass by reference
a=45 b=35

pass-by-value of the reference in java

Refer to this blog
Java is always pass-by-value. The difficult thing to understand is that Java passes objects as references and those references are passed by value. In other words, java is more like passing the const pointer of the object.

Sample 1:

public static void main( String[] args ){
    Dog aDog = new Dog("Max");
    foo(aDog);
    if( aDog.getName().equals("Max") ){ //true
        System.out.println( "Java passes by value." );
    }else if( aDog.getName().equals("Fifi") ){
        System.out.println( "Java passes by reference." );
    }
}

public static void foo(Dog d) {
  d.getName().equals("Max"); // true
  d = new Dog("Fifi");
  d.getName().equals("Fifi"); // true
}

In this example aDog.getName() will still return “Max”. The value aDog within main is not overwritten in the function foo with the Dog “Fifi” as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return “Fifi” after the call to foo.

Dog aDog = new Dog("Max");
foo(aDog);
aDog.getName().equals("Fifi"); // true

public void foo(Dog d) {
  d.getName().equals("Max"); // true
  d.setName("Fifi");
}