why 1 + + "1" === 2; +"1" + + "1" === 2 and "1" + "1" === "11" in javascript [duplicate]

Question

Does the addition operator before char in javascript convert them to number?

1 + + "1" === 2; 
+"1" + + "1" === 2;
"1" + "1" ===  "11"

Earlier question doesn't explain why it's happening only tells us various ways of converting strings to number and vice versa.

Answer 1

1 + + "1" === 2; 

Unary operator + has higher precedence, so +"1" will be evaluated first, converting "1" into integer of value of 1, so it will become

1+1 === 2

The second line

+"1" + + "1" === 2;

This is similar. Unary operator + has higher precedence, so both +"1" will be evaluate to positive integer value of 1.

"1" + "1" ===  "11"

Because in JavaScript + is also string concatenation operator if both operands are string, this will concat both strings.

More information https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Operators/Arithmetic_Operators

Update: @torazaburo is correct.

• The evaluation of +"1" in code here has nothing to do with operator precedence.

• String concatenation will happen if either operand is string.

Answer 2

In JS there is implicit type coercion. For instance, the binary plus operator coerces a number into string if the other operand is a string. "1" + 2 === "12" or 3 + "4" === "34".

However when used with a single operand it works backwards and +"1" coerces the "1" string to a number 1. Just like parseInt("1").

So +"1" + 3 === 4

Answer 3

What you are looking for is the unary plus. Lets analyze the first 2 cases:

The unary plus operator precedes its operand and evaluates to its operand but attempts to convert it into a number, if it isn't already.

1 + + "1" === 2

+ "1" part is converting the string "1" to a number 1. The number 1 is then added to the other number 1 resulting in 2 === 2

+ "1" + + "1" === 2

Same as above, but instead there are 2 unary operation now. So both string "1" are converted to number 1 resulting in 2 === 2