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As far as I understand, short-circuiting with the logical AND && operator works like the following:
Assuming I have the expressions a and b then a && b is the same as a ? b : a since
if a is truthy then the result will be b and
if a is falsy then the result will be a (without even trying to resolve b)
That being the case why is the following (demo) code throwing a SyntaxError:
var add = function(a,b) {
b && return a+b; // if(b) return a+b
...
}
Is there a way to short circuit with a return statement?
439
Short circuiting is an alternative way of using the logical AND or OR operators (& or |) e.g. a non short-circuit OR if(false | true) { } The first condition and second condition are both evaluated even if false is not true (which it always is). However it is was written as a short-circuit OR: if(false || true) { }
A return statement ends the execution of a function, and returns control to the calling function. Execution resumes in the calling function at the point immediately following the call. A return statement can return a value to the calling function. For more information, see Return type.
Short-circuit evaluationThe logical AND expression is a short-circuit operator. As each operand is converted to a boolean, if the result of one conversion is found to be false , the AND operator stops and returns the original value of that falsy operand; it does not evaluate any of the remaining operands.
The Python or operator is short-circuiting When evaluating an expression that involves the or operator, Python can sometimes determine the result without evaluating all the operands. This is called short-circuit evaluation or lazy evaluation. If x is truthy, then the or operator returns x . Otherwise, it returns y .
The && binary operator needs both parts to be expressions.
return something is a statement but not an expression (it doesn't produce a value, as a value wouldn't be useful when the function ends).
Just use
if (b) return a+b;
with the added benefit of an easier to read code.
Read more :
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