why does it always executing the else block of code? [duplicate]

Question

I was trying to make an isNaN() function that checks the user's input if its a NaN or not, but it always skips the the first condition whatever the user's input is.

function isNaN() {
    let value = prompt("enter value");

    if (typeof value === "number") {
    console.log("value is number");
    }else {
    console.log("NaN");
    }
} isNaN();

I made the program to checks the user's input data type by typeof operator.

prompt() always returns a string or null. Even if the string looks numeric, the type is still string. — Barmar
– Barmar, Commented Aug 9, 2024 at 19:23
Use parseFloat() to try to convert the input to a number. It will return NaN if it can't parse it. — Barmar
– Barmar, Commented Aug 9, 2024 at 19:24
This would be a good time to learn to debug unexpected output. Also note that potentially overwriting the default isNaN() implementation, intentionally or not, is not a good idea. — Rory McCrossan
– Rory McCrossan, Commented Aug 9, 2024 at 19:24
Wouldn't it have been simple to just do console.log(typeof value) to see why the condition isn't working as expected? ericlippert.com/2014/03/05/how-to-debug-small-programs — Barmar
– Barmar, Commented Aug 9, 2024 at 19:29

Michael Nyland · Accepted Answer · 2024-08-09 19:35:41Z

0

Try this:

    let value = prompt("enter value");
    console.log("Value entered:" + value);
    console.log("Type of value: " + typeof value);

    if (!isNaN(parseFloat(value))) {
    console.log("String contains a number");
    }else {
    console.log("NaN");
    }

answered Aug 9, 2024 at 19:35

Michael Nyland

16

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