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题目如下
var a = ?; if(a == 1 && a == 2 && a == 3){ conso.log(1); }
The text was updated successfully, but these errors were encountered:
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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题目如下
The text was updated successfully, but these errors were encountered: