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题目如下
var a = ?;
if(a == 1 && a == 2 && a == 3){
conso.log(1);
}
答案解析 因为==会进行隐式类型转换 所以我们重写toString方法就可以了
var a = {
i: 1,
toString() {
return a.i++;
}
}
if( a == 1 && a == 2 && a == 3 ) {
console.log(1);
}
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