Asked  6 Months ago    Answers:  5   Viewed   43 times

I know that in JavaScript you can do:

var oneOrTheOther = someOtherVar || "these are not the droids you are looking for...";

where the variable oneOrTheOther will take on the value of the first expression if it is not null, undefined, or false. In which case it gets assigned to the value of the second statement.

However, what does the variable oneOrTheOther get assigned to when we use the logical AND operator?

var oneOrTheOther = someOtherVar && "some string";

What would happen when someOtherVar is non-false?
What would happen when someOtherVar is false?

Just learning JavaScript and I'm curious as to what would happen with assignment in conjunction with the AND operator.



Basically, the Logical AND operator (&&), will return the value of the second operand if the first is truthy, and it will return the value of the first operand if it is by itself falsy, for example:

true && "foo"; // "foo"
NaN && "anything"; // NaN
0 && "anything";   // 0

Note that falsy values are those that coerce to false when used in boolean context, they are null, undefined, 0, NaN, an empty string, and of course false, anything else coerces to true.

Tuesday, June 1, 2021
answered 6 Months ago
  • ||expr1 || expr2 (Logical OR)

    Returns expr1 if it can be converted to true; otherwise, returns expr2. Thus, when used with Boolean values, || returns true if either operand is true; if both are false, returns false..

  • &&expr1 && expr2 (Logical AND)

    Returns expr1 if it can be converted to false; otherwise, returns expr2. Thus, when used with Boolean values, && returns true if both operands are true; otherwise, returns false.

All values in Javascript are either "truthy" or "falsy".
The following values are equivalent to false in conditional statements:

  • false
  • null
  • undefined
  • The empty string "" ( '')
  • The number 0
  • The number NaN

All other values are equivalent to true.

So... var test = 0 || -1 ; returns -1.

If it was var test = 0 || false || undefined || "" || 2 || -1 it would return 2

Logical operator on MDN

Wednesday, June 9, 2021
answered 6 Months ago

Those are not assignment operators. Those are default arguments for the function.

A function can have one or more default arguments, meaning that if, at the calling point, no argument is provided, the default is used.

void foo(int x = 10) { std::cout << x << std::endl; }

int main()
  foo(5); // will print 5
  foo(); // will print 10, because no argument was provided

In the example code you posted, the ListNode constructor has two parameters with default arguments. The first default argument is Object(), which simply calls the default constructor for Object. This means that if no Object instance is passed to the ListNode constructor, a default of Object() will be used, which just means a default-constructed Object.

See also:
Advantage of using default function parameter
Default value of function parameter

Thursday, July 29, 2021
answered 4 Months ago

Does the compiler generated assignment operator guard against self assignment?

No, it does not. It merely performs a member-wise copy, where each member is copied by its own assignment operator (which may also be programmer-declared or compiler-generated).

Do I always need to protect against self-assignment even when the class members aren't of pointer type?

No, you do not if all of your class's attributes (and therefore theirs) are POD-types.

When writing your own assignment operators you may wish to check for self-assignment if you want to future-proof your class, even if they don't contain any pointers, et cetera. Also consider the copy-and-swap idiom.

Sunday, August 15, 2021
answered 4 Months ago

My programming ruby book (2nd edition) also lists unary operators as having higher precedence than assignment.

The unary operator IS being given highest precedence. The reason the line is parsed as ~ (a = 1) is because decomposing the line into valid syntax is of higher precedence than anything else, including using the simple variable 'a' as the expression the unary operator operates on.

If the ruby parser could have made something valid of the rest of the line, it would have used (~ a), but there is no valid rule than matches = something, only lvalue '=' rvalue.

You can regard "valid syntax" as the top priority, then simple values, constant and variable names and then the standard operators under that.

Wednesday, September 29, 2021
answered 2 Months ago
Only authorized users can answer the question. Please sign in first, or register a free account.
Not the answer you're looking for? Browse other questions tagged :