Trying to add three numbers in javascript using functions but its not adding them instead it just writes them as one number
function numinput(a,b,c,res){
a = prompt("Enter first number");
b = prompt("Enter second number");
c = prompt("Enter third number");
res = a + b + c ;
alert (res);
}
numinput();
Convert the values to numbers using
parseInt
. Here's a working solution.
function numinput(a,b,c,res){
a = parseInt(prompt("Enter first number"), 10);
b = parseInt(prompt("Enter second number"), 10);
c = parseInt(prompt("Enter third number"), 10);
res = a + b + c ;
alert (res);
}
numinput();
prompt returns a string. You need to first convert the string to number otherwise you are concatenating strings: '5' + '7' === '57'
Here are some ways to achieve this:
1 - Using Number
Number('5');
2 - Using parseInt or parseFloat
parseInt('20', 10);
parseFloat('5.5');
3 - Unary + operator as other answers explained
+'5'
Working demo:
function numinput() {
var a = prompt("Enter first number"),
b = prompt("Enter second number"),
c = prompt("Enter third number"),
res = Number(a) + Number(b) + Number(c);
alert(res);
}
numinput();
The each user entry is typeof string, it's being concatenated into one whole string. If you want to add every element as a Math operation, parse the entries into numbers, using + sign in front of variables or parse it using parseInt function.
function numinput(a, b, c, res) {
a = prompt("Enter first number");
b = prompt("Enter second number");
c = prompt("Enter third number");
res = +a + +b + +c;
alert(res);
}
numinput();
You need to convert every value, which is a string, to a number with an unary +.
Then I suggest to move the variables declaration into the function and not inside of the parameters of the function, because you do not need them in this way and you assign the values inside of the function.
function numinput() {
var a = +prompt("Enter first number"),
b = +prompt("Enter second number"),
c = +prompt("Enter third number"),
res = a + b + c;
alert(res);
}
numinput();
const add = function(a, b, c) {
return a + b + c;
}
const result = add(4,5,6);
console.log(result);
