What is the reason behind javaScript not supporting double comparison? For example, in
64 < str.charCodeAt(i) && str.charCodeAt(i)<=7764 < str.charCodeAt(i)<=77?
What is the reason behind javaScript not supporting double comparison? For example, in
64 < str.charCodeAt(i) && str.charCodeAt(i)<=7764 < str.charCodeAt(i)<=77?
Yes, it is possible to write that code, but the outcome may not be what you are expecting.
64 < str.charCodeAt(i)<=77
When this line of code is interpreted, it actually translates to:
(64 < str.charCodeAt(i)) <= 77
This means that the first comparison will occur with the < operator, resulting in a boolean value. Then, this boolean value will be converted to a number (either zero or one) and compared using the <= operator.
The reason why JavaScript behaves this way can be attributed to its origins in the C language family. When designing JavaScript, the creators likely wanted to maintain familiarity and consistency for programmers already accustomed to C programming. On the other hand, Java differs from this behavior due to its use of actual Boolean types and stricter rules against implicit type conversions.
While this code may seem correct, it does not behave as you might anticipate.
The expressions True == 1 and False == 0 both evaluate to true. Therefore, true > x is true for any value of x less than 1, and false > x is true for any value of x less than 0.
As a result, a "double comparison" like:
1 < 2 < 3
is interpreted as:
true < 3
which essentially simplifies to:
1 < 3
and returns true in this scenario.
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