Comments (2)
wangfpp
commented on July 17, 2024
obj2.a 等于3的啊
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jasonGeng88
commented on July 17, 2024
解释内容有误，已修正。
解释如下：
先获取obj2.obj1 -> 通过引用获取到obj1对象，再访问 obj1.foo -> 最后执行foo函数调用
这里调用链不只一层，存在obj1、obj2两个对象，那么隐式绑定具体会绑哪个对象。这里原则是获取最后一层调用的上下文对象，即obj1，所以结果显然是4（obj1.a）。
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