这里有一个例子:

#include <iostream>

void by_val(int arg) { arg += 2; }
void by_ref(int&arg) { arg += 2; }

int main()
{
    int x = 0;
    by_val(x); std::cout << x << std::endl;  // prints 0
    by_ref(x); std::cout << x << std::endl;  // prints 2

    int y = 0;
    by_ref(y); std::cout << y << std::endl;  // prints 2
    by_val(y); std::cout << y << std::endl;  // prints 2
}

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

下面是一个例子，演示了通过值-指针值-引用传递之间的区别:

void swap_by_value(int a, int b){
    int temp;

    temp = a;
    a = b;
    b = temp;
}   
void swap_by_pointer(int *a, int *b){
    int temp;

    temp = *a;
    *a = *b;
    *b = temp;
}    
void swap_by_reference(int &a, int &b){
    int temp;

    temp = a;
    a = b;
    b = temp;
}

int main(void){
    int arg1 = 1, arg2 = 2;

    swap_by_value(arg1, arg2);
    cout << arg1 << " " << arg2 << endl;    //prints 1 2

    swap_by_pointer(&arg1, &arg2);
    cout << arg1 << " " << arg2 << endl;    //prints 2 1

    arg1 = 1;                               //reset values
    arg2 = 2;
    swap_by_reference(arg1, arg2);
    cout << arg1 << " " << arg2 << endl;    //prints 2 1
}

“参照传递”方法有一个重要的局限性。如果参数声明为引用传递(因此前面有&号)，其对应的实际参数必须是变量。

引用“按值传递”形式参数的实际参数一般可以是表达式，因此它不仅可以使用变量，还可以使用文字甚至函数调用的结果。

该函数不能将值放在变量以外的东西中。它不能将新值赋给文字或强制表达式更改其结果。

PS:你也可以在当前的线程中检查Dylan Beattie的答案，它用简单的语言解释了它。

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)。

看看这张照片:

在第一种情况下(通过引用传递)，当变量在函数内部设置或更改时，外部变量也会更改。

但在第二种情况下(按值传递)，改变函数内部的变量不会对外部变量产生任何影响。

要阅读这篇文章，请参见这个链接。

最简单的方法是在Excel文件中获取。举个例子，有两个数字，5和2分别在A1和B1单元格中，你想在第三个单元格中求出它们的和，假设是A2。

有两种方法可以做到这一点。

通过在单元格中输入= 5 + 2将它们的值传递给单元格A2。在这种情况下，如果单元格A1或B1的值发生变化，则A2中的总和保持不变。 或者通过输入= A1 + B1将单元格A1和B1的“引用”传递给单元格A2。在这种情况下，如果单元格A1或B1的值发生变化，A2中的总和也会发生变化。