“按值传递”将发送存储在指定变量中的数据的副本，“按引用传递”将发送到变量本身的直接链接。

因此，如果你通过引用传递一个变量，然后改变你传递给它的块中的变量，原始变量将被改变。如果只是按值传递，原始变量将不能被传递到的块所改变，但您将获得调用时它所包含的任何内容的副本。

1. 按值传递/按值调用

   void printvalue(int x) 
   {
       x = x + 1 ;
       cout << x ;  // 6
   }

   int x = 5;
   printvalue(x);
   cout << x;    // 5

在按值调用中，当你将一个值传递给printvalue(x)时，即参数5，它被复制为void printvalue(int x)。现在，我们有两个不同的值5和复制的值5，这两个值存储在不同的内存位置。因此，如果你在void printvalue(int x)内做了任何更改，它不会反射回实参。

2. 引用传递/引用调用

   void printvalue(int &x) 
   {
      x = x + 1 ;
      cout << x ; // 6
   }

   int x = 5;
   printvalue(x);
   cout << x;   // 6

在引用调用中，只有一个区别。我们使用&，即地址操作符。通过做 Void printvalue(int &x)我们引用的是x的地址，这告诉我们它指向相同的位置。因此，在函数内部所做的任何更改都会在函数外部反映出来。

既然你来了，你也应该知道……

3.按指针传递/按地址调用

   void printvalue(int* x) 
   {
      *x = *x + 1 ;
      cout << *x ; // 6
   }

   int x = 5;
   printvalue(&x);
   cout << x;   // 6

在传递地址中，指针int* x保存传递给它的地址printvalue(&x)。因此，在函数内部所做的任何更改都会在函数外部反映出来。

通过值传递-函数复制变量并使用副本(因此它不会改变原始变量中的任何内容)

引用传递——函数使用原始变量。如果你改变了另一个函数中的变量，它也会改变原来的变量。

示例(复制并使用/自己尝试一下):

#include <iostream>

using namespace std;

void funct1(int a) // Pass-by-value
{
    a = 6; // Now "a" is 6 only in funct1, but not in main or anywhere else
}

void funct2(int &a)  // Pass-by-reference
{
    a = 7; // Now "a" is 7 both in funct2, main and everywhere else it'll be used
}

int main()
{
    int a = 5;

    funct1(a);
    cout << endl << "A is currently " << a << endl << endl; // Will output 5
    funct2(a);
    cout << endl << "A is currently " << a << endl << endl; // Will output 7

    return 0;
}

保持简单，伙计们。成堆的文字是个坏习惯。

It's a way how to pass arguments to functions. Passing by reference means the called functions' parameter will be the same as the callers' passed argument (not the value, but the identity - the variable itself). Pass by value means the called functions' parameter will be a copy of the callers' passed argument. The value will be the same, but the identity - the variable - is different. Thus changes to a parameter done by the called function in one case changes the argument passed and in the other case just changes the value of the parameter in the called function (which is only a copy). In a quick hurry:

Java only supports pass by value. Always copies arguments, even though when copying a reference to an object, the parameter in the called function will point to the same object and changes to that object will be see in the caller. Since this can be confusing, here is what Jon Skeet has to say about this. C# supports pass by value and pass by reference (keyword ref used at caller and called function). Jon Skeet also has a nice explanation of this here. C++ supports pass by value and pass by reference (reference parameter type used at called function). You will find an explanation of this below.

代码

因为我的语言是c++，我将在这里使用它

// passes a pointer (called reference in java) to an integer
void call_by_value(int *p) { // :1
    p = NULL;
}

// passes an integer
void call_by_value(int p) { // :2
    p = 42;
}

// passes an integer by reference
void call_by_reference(int & p) { // :3
    p = 42;
}

// this is the java style of passing references. NULL is called "null" there.
void call_by_value_special(int *p) { // :4
    *p = 10; // changes what p points to ("what p references" in java)
    // only changes the value of the parameter, but *not* of 
    // the argument passed by the caller. thus it's pass-by-value:
    p = NULL;
}

int main() {
    int value = 10;
    int * pointer = &value;

    call_by_value(pointer); // :1
    assert(pointer == &value); // pointer was copied

    call_by_value(value); // :2
    assert(value == 10); // value was copied

    call_by_reference(value); // :3
    assert(value == 42); // value was passed by reference

    call_by_value_special(pointer); // :4
    // pointer was copied but what pointer references was changed.
    assert(value == 10 && pointer == &value);
}

Java中的一个例子不会有什么坏处:

class Example {
    int value = 0;

    // similar to :4 case in the c++ example
    static void accept_reference(Example e) { // :1
        e.value++; // will change the referenced object
        e = null; // will only change the parameter
    }

    // similar to the :2 case in the c++ example
    static void accept_primitive(int v) { // :2
        v++; // will only change the parameter
    }        

    public static void main(String... args) {
        int value = 0;
        Example ref = new Example(); // reference

        // note what we pass is the reference, not the object. we can't 
        // pass objects. The reference is copied (pass-by-value).
        accept_reference(ref); // :1
        assert ref != null && ref.value == 1;

        // the primitive int variable is copied
        accept_primitive(value); // :2
        assert value == 0;
    }
}

维基百科

http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_value

http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_reference

这家伙说得很准:

http://javadude.com/articles/passbyvalue.htm

例子:

class Dog 
{ 
public:
    barkAt( const std::string& pOtherDog ); // const reference
    barkAt( std::string pOtherDog ); // value
};

Const &通常是最好的。你不会受到建造和破坏的惩罚。如果引用不是const，你的接口暗示它将改变传入的数据。

问题是“vs”。

没有人指出一个重要的点。在传递值时，会占用额外的内存来存储传递的变量值。

在传递引用时，值不会占用额外的内存(在某些情况下内存是有效的)。