I'm trying to understand switch behaviour when it deals with false.

let grade = 65;

switch(false){
  case grade >= 90:
    console.log(grade);
    console.log("You did great!");
    break;
  case grade >= 80:
    console.log("You did good!");
    break;
  default:
    console.log(grade, "is not a letter grade!");

I don't understand why grade will always hit first case in the code above

I was expecting none of the case being fulfilled because of switch(false), and there should be no console log printed.

CodePudding user response：

swtich compares the expression in the case with the value passed to switch.

(grade >= 90) === false

---

I strongly recommend only using simple values in cases and using if / else for more complex logic like you have here. Putting expressions in cases is unintuative.

CodePudding user response：

You need to check against the opposite of the expression, because of the double negation.

switch (false) {
    case !(grade >= 90): // code
}

vs

switch (true) {
    case grade >= 90: // code
}

CodePudding user response：

From MDN:

The switch statement evaluates an expression, matching the expression's value against a series of case clauses, and executes statements after the first case clause with a matching value, until a break statement is encountered. The default clause of a switch statement will be jumped to if no case matches the expression's value.

More details here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/switch

CodePudding user response：

It's because, it compares passed value (false) with case equation (which is also false in first case), so false === false is true (Yes it's ===, not ==), so it's going into the first case.