当我们在函数中将obj2分配给obj1时,按引用传递不起作用,在函数外部它不起作用并且obj1保留其原始值,但是为什么呢?
let obj1 = {
value: 'a'
}
let obj2 = {
value: 'b'
}
obj3 = obj2;
function change(obj1, obj2) {
obj1 = obj2
obj2.value = 'c'
}
change(obj1, obj2);
console.log(obj1.value)Run Code Online (Sandbox Code Playgroud)
javascript ×1