CODEWITHSUNDEEP
javascriptarraysfunction
NoChance
NoChance

Reputation: 5752

Why console.log function returns undefined?

function stringifyObj(parmObj){
    s="";
    Object.getOwnPropertyNames(parmObj).forEach
    (
        function (val, idx, array) {
          s+=val + ' -> ' + parmObj[val]+"\n";
        }
    )
    return s;
}

var arrayOfObjects = [
    { name: 'Edward', value: 21 },
    { name: 'Sharpe', value: 37 }
  ];

console.log(
    arrayOfObjects.forEach(function(parmArrItem)
        {
            const p=stringifyObj(parmArrItem);
            console.log(p); 
        }
));

In the code below, 2 objects are displayed fine, but after that I get undefined displayed at the end of the run. Where does the undefined come from? Thanks.

Upvotes: 2

Views: 6081

Answers (1)

Mihai Alexandru-Ionut
Mihai Alexandru-Ionut

Reputation: 48357

arrayOfObjects.forEach returns nothing.

So, when you use console.log() for a void function that's you received undefined.

forEach method only executes a callback provided function for every item from an array.

With the other words, the console prints the result of evaluating an expression.

console.log() is undefined since your function or expression not explicitly return something.

function stringifyObj(parmObj){
    s="";
    Object.getOwnPropertyNames(parmObj).forEach
    (
        function (val, idx, array) {
          s+=val + ' -> ' + parmObj[val]+"\n";
        }
    )
    return s;
}

var arrayOfObjects = [
    { name: 'Edward', value: 21 },
    { name: 'Sharpe', value: 37 }
  ];


arrayOfObjects.forEach(function(parmArrItem)
        {
            const p=stringifyObj(parmArrItem);
            console.log(p); 
        }
);

Upvotes: 7

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