Simple js FOR loop returning 'undefined'

问题

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

function translate2(x){
  var newStr;
  x = "Hello World";
  for(i=0; i<x.length; i++) {
    newStr+=x.charAt(i);
  }
  console.log(newStr);
}

回答1:

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;
}
```

回答2:

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.

回答3:

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);
}

回答4:

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!

来源：https://stackoverflow.com/questions/28665160/simple-js-for-loop-returning-undefined

标签

javascript

variables

for-loop