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When I type in browser's console:
console.log(2 && 3)
it results always with second number (in this case 3):
3
Can someone explain me why?
If the left hand side of && evaluates as a false value, the whole expression evaluates as the left hand side.
Otherwise it evaluates as the right hand side.
2 is a true value, so 2 && 3 is 3.
For comparison, try console.log(0 && 1) and console.log(false && "something").
The && logical operator will return the last value if all other values are truthy else it will return the first non truthy value.
So in your case since 2 is truthy then it will evaluate 3 and since that is truthy it will be returned.
Same way 2 && 0 && 4 will return 0 as it is a non truthy value.
Logical operators are typically used with Boolean (logical) values. When they are, they return a Boolean value. However, the && and || operators actually return the value of one of the specified operands, so if these operators are used with non-Boolean values, they may return a non-Boolean value.
&& has to evaluate all expressions. 2 && 3 will first evaluate the “truthiness” of 2 which is a true value but then it has to evaluate 3 as well. That last evaluated value is then returned. If at least one value is non-truthy then the first such value is returned instead.
|| on the other hand returns the first truthy expression or the last non-truthy if there are no truthy expressions.
The reason why && returns the last possible value is that it simply has to check all expressions in order to return a result. || doesn’t have to do that. If the first expression for || is true it ignores all further ones. Likewise if the first expression for && is false it ignores all further ones (see short-circuiting in logical operators).
来源：https://stackoverflow.com/questions/30252983/why-does-2-3-results-in-3-javascript