2/4 Somehow doesn't work


#1
var bob = {
    firstName: "Bob",
    lastName: "Jones",
    phoneNumber: "(650) 777-7777",
    email: "bob.jones@example.com"
};

var mary = {
    firstName: "Mary",
    lastName: "Johnson",
    phoneNumber: "(650) 888-8888",
    email: "mary.johnson@example.com"
};

var contacts = [bob,mary];

// printPerson added here
var printPerson=function(person){
  return person.firstName+" "+person.lastName;
    
    };

printPerson(contacts[0]);
printPerson(contacts[1]);

Tried this code on a webpage using alert instead of return and it worked, tried refreshing and still the same problem, it only shows "Mary Johnson"


#2

The output you are seeing is the response from the command line interpreter to the last executed statement,

return person.firstName + " " + person.lastName;

What we see is the return value of printPerson(contacts[1]);. To fix this, have your function log the output, rather than return it.

console.log(person.firstName + " " + person.lastName);

It works, but it seems wrong
#3

++++ function with 1 parameter using return-statement

var myFunction = function( param1 ) {
       //Begin of FUNCTION-BODY
       //myFunction =function= has 1 PARAMETER param1
       //this param1 PARAMETER is used as a -local- VARIABLE
       //throughout the FUNCTION-BODY

      return param1;

      //End of FUNCTION-BODY
      };

you have defined a myFunction function
which takes 1 parameter param1
this param1 parameter is used
as a variable throughout the FUNCTION-BODY.

If you want to call/execute this myFunction function
and this myFunction function was defined
as having 1 parameter param1
you will have to provide 1 argument
in our case a "number VALUE" 4
myFunction( 4 );

some quotes from the outer-world:

**argument is the value/variable/reference being passed in,
parameter is the receiving variable used within the function/block**

OR

**"parameters" are called "formal parameters",
while "arguments" are called "actual parameters".**


#4

As you are using the return-statement
you might try using
console.log( printPerson(contacts[0]) );
console.log( printPerson(contacts[1]) );


#5

Thanks for the replies i understand better now.