两者有什么区别
引用传递的参数 参数通过value?
你能给我举几个例子吗?
当前回答
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
其他回答
如果你不想在将原始变量传递给函数后改变它的值，那么函数应该构造一个“按值传递”参数。
然后函数将只有值，而没有传入变量的地址。如果没有变量的地址，函数内部的代码就不能改变从函数外部看到的变量值。
但是如果你想要赋予函数从外部看到的改变变量值的能力，你需要使用引用传递。因为值和地址(引用)都是传递进来的，并且在函数内部可用。
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
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)。
简而言之，通过值传递的是它是什么，通过引用传递的是它在哪里。
如果你的值是VAR1 = "Happy Guy!"，你只会看到"Happy Guy!"如果VAR1变成了“Happy Gal!”，你就不会知道了。如果它是通过引用传递的，并且VAR1发生了变化，那么您就会这样做。
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。
