Not sure what I'm doing wrong here; the variable newStr should just return "Hello World", but I'm getting this instead:
"undefinedHello World"
undefined
JS
function translate2(x){
var newStr;
x = "Hello World";
for(i=0; i<x.length; i++) {
newStr+=x.charAt(i);
}
console.log(newStr);
}
In JavaScript, if a variable is not initialized explicitly, it will by default have undefined. That is not a string but a primitive type of the Language. You can check that by printing it
var newStr;
console.log(newStr);
// undefined
console.log(newStr + "thefourtheye");
// undefinedthefourtheye
So, just initialize the variable with an empty string, like this
var newStr = '';
Also, note that, in this line
for(i=0; i < x.length; i++) {
i has never been declared before. So, a new global variable i will be created. You may not want that. So, just use var keyword to declare the variable scoped to the current function, like this
for (var i = 0; i < x.length; i++) {
Apart from that, translate2 is a function and when it is invoked, one would expect it to return something. But you are not returning anything explicitly. So, again, JavaScript, by default, returns undefined. That is why you are getting the second undefined in the question. To fix that, use return statement like this
function translate2(x) {
var newStr = "";
for (var i = 0; i < x.length; i++) {
newStr += x.charAt(i);
}
return newStr;
}
console.log(translate2(..)), try simply translate2(..). The console.log function call doesn't return anything, so undefined will be returned by default. That is why you are getting an extra undefined in the console.translate2()You should first initialize the variable newStr.
var newStr = '';
Otherwise, newStr will be undefined and undefined + "asa" = "undefinedasa" in javascript. If you don't know what is undefined, check this out.
newStr is undefined. Add
var newStr = '';
So that you have
function translate2(x){
var newStr='';
x = "Hello World";
for(i=0; i<x.length; i++) {
newStr+=x.charAt(i);
}
console.log(newStr);
}
The above answers are not correct. The console.log() will run before the loop finishes and that is why you get undefiend.
You can find your answer here.
you have to think sync like this piece of code:
function delay() {
return new Promise(resolve => setTimeout(resolve, 300));
}
async function delayedLog(item) {
// notice that we can await a function
// that returns a promise
await delay();
console.log(item);
}
async function processArray(array) {
for (const item of array) {
await delayedLog(item);
}
console.log('Done!');
}
processArray([1, 2, 3]);
this will give you 1,2,3,done which means the console.log is happening at the end of loop!
var newStr = '';undefined + 'qwe'==='undefinedqwe'