基元类型（数字、字符串等）是通过值传递的，但对象是未知的，因为它们既可以通过值传递（如果我们认为持有对象的变量实际上是对对象的引用），也可以通过引用传递（当我们认为对象的变量持有对象本身）。
虽然最后这并不重要，但我想知道通过约定来表达论点的正确方式是什么。是否有JavaScript规范的摘录，该规范定义了与此相关的语义？
当前回答
在JavaScript中向函数传递参数类似于传递参数（按C中的指针值）：
/*
The following C program demonstrates how arguments
to JavaScript functions are passed in a way analogous
to pass-by-pointer-value in C. The original JavaScript
test case by @Shog9 follows with the translation of
the code into C. This should make things clear to
those transitioning from C to JavaScript.
function changeStuff(num, obj1, obj2)
{
num = num * 10;
obj1.item = "changed";
obj2 = {item: "changed"};
}
var num = 10;
var obj1 = {item: "unchanged"};
var obj2 = {item: "unchanged"};
changeStuff(num, obj1, obj2);
console.log(num);
console.log(obj1.item);
console.log(obj2.item);
This produces the output:
10
changed
unchanged
*/
#include <stdio.h>
#include <stdlib.h>
struct obj {
char *item;
};
void changeStuff(int *num, struct obj *obj1, struct obj *obj2)
{
// make pointer point to a new memory location
// holding the new integer value
int *old_num = num;
num = malloc(sizeof(int));
*num = *old_num * 10;
// make property of structure pointed to by pointer
// point to the new value
obj1->item = "changed";
// make pointer point to a new memory location
// holding the new structure value
obj2 = malloc(sizeof(struct obj));
obj2->item = "changed";
free(num); // end of scope
free(obj2); // end of scope
}
int num = 10;
struct obj obj1 = { "unchanged" };
struct obj obj2 = { "unchanged" };
int main()
{
// pass pointers by value: the pointers
// will be copied into the argument list
// of the called function and the copied
// pointers will point to the same values
// as the original pointers
changeStuff(&num, &obj1, &obj2);
printf("%d\n", num);
puts(obj1.item);
puts(obj2.item);
return 0;
}
其他回答
在低级语言中，如果要通过引用传递变量，则必须在创建函数时使用特定语法：
int myAge = 14;
increaseAgeByRef(myAge);
function increaseAgeByRef(int &age) {
*age = *age + 1;
}
&age是对myAge的引用，但如果您需要该值，则必须使用*age转换引用。
JavaScript是一种高级语言，可以为您进行这种转换。
因此，尽管对象是通过引用传递的，但语言将引用参数转换为值。您不需要在函数定义上使用&来通过引用传递它，也不需要在功能体上使用*来将引用转换为值，JavaScript会为您执行此操作。
这就是为什么当您试图通过替换函数中的对象的值（即age={value:5}）来更改该对象时，更改不会持久，但如果您更改了它的财产（即age.value=5），更改就会持久。
了解更多信息
这里有一些关于JavaScript中使用术语“通过引用传递”的讨论，但要回答您的问题：
对象通过引用自动传递，无需特别声明
（摘自上述文章。）
变量不“保存”对象；它有一个参考。您可以将该引用分配给另一个变量，现在两者都引用同一个对象。它总是按值传递（即使该值是引用…）。
无法更改作为参数传递的变量所持有的值，如果JavaScript支持通过引用传递，这是可能的。
在JavaScript中向函数传递参数类似于传递参数（按C中的指针值）：
/*
The following C program demonstrates how arguments
to JavaScript functions are passed in a way analogous
to pass-by-pointer-value in C. The original JavaScript
test case by @Shog9 follows with the translation of
the code into C. This should make things clear to
those transitioning from C to JavaScript.
function changeStuff(num, obj1, obj2)
{
num = num * 10;
obj1.item = "changed";
obj2 = {item: "changed"};
}
var num = 10;
var obj1 = {item: "unchanged"};
var obj2 = {item: "unchanged"};
changeStuff(num, obj1, obj2);
console.log(num);
console.log(obj1.item);
console.log(obj2.item);
This produces the output:
10
changed
unchanged
*/
#include <stdio.h>
#include <stdlib.h>
struct obj {
char *item;
};
void changeStuff(int *num, struct obj *obj1, struct obj *obj2)
{
// make pointer point to a new memory location
// holding the new integer value
int *old_num = num;
num = malloc(sizeof(int));
*num = *old_num * 10;
// make property of structure pointed to by pointer
// point to the new value
obj1->item = "changed";
// make pointer point to a new memory location
// holding the new structure value
obj2 = malloc(sizeof(struct obj));
obj2->item = "changed";
free(num); // end of scope
free(obj2); // end of scope
}
int num = 10;
struct obj obj1 = { "unchanged" };
struct obj obj2 = { "unchanged" };
int main()
{
// pass pointers by value: the pointers
// will be copied into the argument list
// of the called function and the copied
// pointers will point to the same values
// as the original pointers
changeStuff(&num, &obj1, &obj2);
printf("%d\n", num);
puts(obj1.item);
puts(obj2.item);
return 0;
}
我发现Undercore.js库的extend方法非常有用，当我想将一个对象作为一个参数传递时，它可以被修改或完全替换。
function replaceOrModify(aObj) {
if (modify) {
aObj.setNewValue('foo');
} else {
var newObj = new MyObject();
// _.extend(destination, *sources)
_.extend(newObj, aObj);
}
}
