JavaScript – Why Does ([] == false) is True but ([] || true) Results in []?

comparisonjavascripttruthiness

Was just doing some testing and I find this odd:

[] == false

Gives true, this makes sense because double equal only compares contents and not type and tries to do type-coercion. But if its comparing contents and returns true, that means [ ] is falsey (if you did [] == true you get false too), which means:

[] || false

Should give false, but it gives [ ], making it truthy? Why?

Another example:

"\n  " == 0

Gives true, but "\n " || false gives "\n "? Is there an explanation for this or its just an oddity.

When I tried this in C, we get:

int x = "\n " == 0;
printf("%d\n", x);

int y = "\n " || 0;
printf("%d\n", y);

Outputs:

0
1

This makes sense, but given C's influence on Javascript, the behaviour is different.

Best Answer

Type conversion is not related to falsy and truthy values.

What is truthy and what is falsy is defined by the ToBoolean function defined in the specs and [] is indeed truthy.

On the other hand, [] == false returns true because of the type conversion that happens during the evaluation of the expression.

The rules of type conversion say that for x == y

If Type(y) is Boolean, return the result of the comparison x == ToNumber(y).

ToNumber results in 0 for false so we're left with the evaluation of [] == 0. According to the same rules

If Type(x) is Object and Type(y) is either String or Number, return the result of the comparison ToPrimitive(x) == y.

ToPrimitive results in an empty string. Now we have "" == 0. Back to our type conversion rules

If Type(x) is String and Type(y) is Number, return the result of the comparison ToNumber(x) == y.

ToNumber results in 0 for "" so the final evaluation is 0 == 0 and that is true!

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