This is my code
function test(number) {
if (number % 15 == 0) {
return "FizzBuzz";
}
if (number % 3 == 0) {
return "Fizz";
}
if (number % 5 == 0) {
return "Buzz";
}
return number.toString();
}
I type test(15).
result print "FizzBuzz"
Why doesn't it print "FizzBuzz" "Fizz" and "Buzz" ?
You can only return once per function. After return it goes back to the line were you called the function.
When you return, the fuction is terminated and the passed value is returned. Any code that follows the return is not executed.
In your case, the function enters the first statement (as the number % 15 == 0 statement returns true), sees a return, returns FizzBuzz, and breaks out of the function.
Because you leave the function with a return write this instead
function test(number) {
var res = "";
if(number % 15 == 0){
res += "FizzBuzz";
}
if(number % 3 == 0){
res += "Fizz";
}
if(number % 5 ==0){
res += "Buzz";
}
return res;
}
It will add the results from the conditions.
@Yehuda correctly explained why it only display FizzBuzz: The execution of a function is stopped when you return.
Here is one of the (many) solutions to get all the modulos: It defines an array which in you can store the modulos. At the end of the code, it returns the array.
function test(number) {
var res = [];
if(number % 15 == 0) {
res.push("FizzBuzz");
}
if(number % 3 == 0) {
res.push("Fizz");
}
if(number % 5 == 0) {
res.push("Buzz");
}
console.log(res);
return res;
}
test(15);If you want to return varius:
function test(number) {
res = '';
if(number % 15 == 0){
res += " FizzBuzz";
}
if(number % 3 == 0){
res += " Fizz";
}
if(number % 5 ==0){
res += " Buzz";
}
return number + " - " + res.toString();
}
returnbreaks out of the functionreturnwhich means: at this point jump out of function and return the specific value