The "fn" console.log comes up before than console.log of the two variables.
My function is that:
function test1(var1, var2, fn){
console.log(var1, var2);
fn();
}
function test2(var3){
console.log(var3 + " it's here");
}
Call:
test1(123, "Hello!", test2("Look") );
You're not passing a function as the third argument, you're calling the function and passing its returned value. It should be:
test1(123, "Hello!", function() { test2("Look"); });
In addition to get the the wrong order of output, you should also be getting an error when you try to call fn(), since fn is undefined.
When you call test1(123, "Hello!", test2("Look") );, this is what's happening:
test2("Look")test1: test1(123, "Hello!", undefined);Basically, test2 is executed before test1 is called, because you pass the return value of the function as a parameter.
To actually pass the function itself, to execute "later", you'll need to wrap it in an anonymous function:
test1(123, "Hello!", function() { test2("Look"); });
code should looks like:
(function(){
function test1(var1, var2, fn){
console.log(var1, var2);
fn("Look");
}
function test2(var3){
console.log(var3 + " its here");
}
test1(123, "Hello!", test2);
})();
BTW "it's here" - ' (single quotation mark) got more power then " (double quotation mark). It should looks like 'it\'s here' or "it\'s here".
and if you want to call the passing function it should looks like @Cerbrus and @Barmar said:
test1(123, "Hello!", function() { test2("Look"); });
test1, you need to know the values of all of its arguments.test2("Look")must be evaluated first.test1.