Invoking a function sharing the same name as a local variable

Question

The code snippet below is causing me some trouble...

var firstPlay = 1;
if (firstPlay == 1) {
    firstPlay();
}

If I remove the if statement and simply have:

 firstPlay();

It works fine, but for some reason it doesn't work with the if statement.

I'm wondering if there's a typo in my code. If not, how should I approach debugging this issue?

Answer 1

Answer №1

When assigning a value to the variable firstPlay, make sure to avoid using numerical values as names for your variables.

Consider using more descriptive variable names like:

var firstPlayTest = 1;
if (firstPlayTest == 1) { firstPlay(); }

Answer 2

When assigning a value to the variable firstPlay, make sure to avoid using numerical values as names for your variables.

Consider using more descriptive variable names like:

var firstPlayTest = 1;
if (firstPlayTest == 1) { firstPlay(); }

Answer 3

Function firstPlay() previously defined is now reassigned with the value 1 in the code snippet below:

        function firstPlay() {alert("firstPlay");}
        alert(firstPlay);
        var firstPlay = 1;
        alert(firstPlay);

The initial alert displays the function definition, whereas the second alert shows the number "1".

Answer 4

Function firstPlay() previously defined is now reassigned with the value 1 in the code snippet below:

        function firstPlay() {alert("firstPlay");}
        alert(firstPlay);
        var firstPlay = 1;
        alert(firstPlay);

The initial alert displays the function definition, whereas the second alert shows the number "1".