Magic variables in JavaScript that will drive you crazy

There is such a variable tt, which satisfies the following code. After the code is executed, errCount=0 and the assertion function never alerts information The code is as follows:

Copy code The code is as follows: var tt=/* Please define tt here */; var errCount=0 //Global variable, used to record the number of assertion declarations in the assert function /* Assertion function If v is false, the function alert("assert error") and accumulates the counter errCount If v is true, do nothing */ function assert(v) { if (!v) { alert("assert error"); errCount; } }

assert((tt || true) == false) assert((tt || false) == false) assert((tt && true ) == true) assert((tt && false) == false) assert((true || tt) == true) assert((tt || true) == false) assert((false || tt ) == false) assert((tt || false) == false) assert((true && tt) == false) assert((tt && true ) == true) assert((false && tt ) == false) assert((tt && false ) == false)

assert((tt ? true : false) == true) assert((tt == false) == true) assert((!tt == tt ) == true) assert((tt '') == "false" ) assert(tt

Tongfa CSDN: Magic Variables in JavaScript This question was a question that one of my colleagues asked me to test after get off work. I thought about it for a long time and tried for a long time. I got the following answer , can meet the requirements of the question. The answer is as follows: var tt=new Object(false); var tt=new Boolean(); var tt=new Boolean(false); Through this question, we can feel the flexibility of js and deepen our understanding of js. I used to often use the following judgments in the code if (a){ alert('ok') } Now it seems that the logical hidden danger of writing this way is very big. I will use the wonderful reply of a netizen in csdn to explain the above problem || is calculated like this: From the first From the beginning, if a meaningful return is encountered, otherwise the last expression is returned (note that it is not necessarily a Boolean value); && is operated like this: starting from the first one, if a meaningless return is encountered, otherwise it is returned The last expression (note the same as above); ! is operated like this: negate the value of the expression (note not the expression). What is meaningless: the following six: 0,null,undefined,"",false,NaN Other than these, it is regarded as meaningful. new Boolean(), new Boolean(false) are the same thing. Since it is an object, it is meaningful, but its value is false, so it can be regarded as "meaningful false". In this way, That can explain all the problems. new Object(false) is also meaningful, its value is also false, but its type is Object, while new Boolean()'s type is Boolean. One more reminder: the two operators || and && do not operate on values, that is, regardless of the value of the expression during the operation, they operate on the expression itself; These two operators How do you evaluate expressions? Answer: It only matters whether the expression is meaningful, regardless of its value. For expressions, there are only 6 meaningless ones; please note here: all objects generated by new method are dynamic objects, and dynamic objects are regarded as meaningful Two more examples: 1. 0||false||new Boolean(false) The operation is as follows: First of all: 0 is a constant, which happens to be one of the meaningless ones, so continue; and false is also one of the meaningless ones, so continue; new Boolean(false) is a dynamic object and is meaningful, so the result of the above operation is new Boolean(false) 2. 0||new Boolean(false)||true What will be the result? Many people will think that the result is true without paying attention, but this is wrong. Let’s talk about the answer first: the result is the same as above First of all: 0 is a constant, which happens to be one of the meaningless ones, so continue; new Boolean (false) is a dynamic object, meaningful; a meaningful expression has been found here, So no further calculations are needed. So the result is the same as above. ====== The same applies to the && operator, so I won’t say more. Off-topic: What’s interesting is that after the answer came out, we also used some unconventional means to answer this question (for entertainment only)

var tt=window[" assert"]=new Function(); This sentence means the following two lines of code function assert(){} function tt(){} javascript allows duplication Define a function, and when executing it, the latter one shall prevail.

Magic variables in JavaScript that will drive you crazy_javascript tips