What is the appropriate method to call a function when a parameter is unnecessary?

Question

What is the appropriate method to call a function when a parameter is unnecessary?

Just to confirm, is this the correct way to call a function when a parameter is not needed?

function load_img(src, alt, el, other_src) {

// check if other_src exists, not null, defined, not empty, etc...

}

//call the function
load_img(src, alt, '#zoom-cover');

Is it better to write:

load_img(src, alt, '#zoom-cover', null);

or

load_img(src, alt, '#zoom-cover', '');

Is there a similar concept like in PHP where we can set a default value for the parameter?

load_img(src, alt, '#zoom-cover', other_src='default value');

Also, how can I ensure within the function that other_src exists, is defined, has a valid value, and is not null or an empty string?

javascript

Answer 1

Answer №1

Everything is okay

display_image(source, description, '#zoom-cover');

To set a default value for the other_source parameter when it is not provided or null, follow these steps:

function display_image(source, description, element, other_source) {
  if (!other_source) {
    other_source = "defaultcontent"
  }
}

Alternatively, you can use this approach to assign the default value to other_source only when the parameter is omitted.

function display_image(source, description, element, other_source) {
  if (typeof(other_source) === "undefined") {
    other_source = "defaultcontent"
  }
}

Answer 2

Everything is okay

display_image(source, description, '#zoom-cover');

To set a default value for the other_source parameter when it is not provided or null, follow these steps:

function display_image(source, description, element, other_source) {
  if (!other_source) {
    other_source = "defaultcontent"
  }
}

Alternatively, you can use this approach to assign the default value to other_source only when the parameter is omitted.

function display_image(source, description, element, other_source) {
  if (typeof(other_source) === "undefined") {
    other_source = "defaultcontent"
  }
}

Answer 3

Answer №2

To remove additional parameters, simply leave them out.

If you wish to set a default value, you can do so by using

other_src = other_src || 'default value';

at the beginning of your function. However, this may not be suitable if the original value could potentially be falsy. In such cases, consider:

if (typeof other_src === 'undefined') {
    other_src = 'default value';
}

Answer 4

To remove additional parameters, simply leave them out.

If you wish to set a default value, you can do so by using

other_src = other_src || 'default value';

at the beginning of your function. However, this may not be suitable if the original value could potentially be falsy. In such cases, consider:

if (typeof other_src === 'undefined') {
    other_src = 'default value';
}

Answer 5

Answer №3

When working with JavaScript, keep in mind that if a parameter is not passed, the local variable will default to being of type 'undefined', allowing you to still call the function without providing all parameters.

load_img(source, alternativeText, '#zoom-cover');

Answer 6

When working with JavaScript, keep in mind that if a parameter is not passed, the local variable will default to being of type 'undefined', allowing you to still call the function without providing all parameters.

load_img(source, alternativeText, '#zoom-cover');

Answer 7

Answer №4

If the last argument is not necessary, you can simply leave it out. For example:

load_img(src, alt, '#zoom-cover');

As for your second inquiry, you could handle it like this:

if(typeof other_src === 'undefined') {
    ...
}

In JavaScript, there are no named arguments like in PHP. However, you can pass an object instead, such as:

load_img(src, alt, '#zoom-cover', {other_src: 'something'});

This approach is commonly seen in jQuery Plugins.

Answer 8

If the last argument is not necessary, you can simply leave it out. For example:

load_img(src, alt, '#zoom-cover');

As for your second inquiry, you could handle it like this:

if(typeof other_src === 'undefined') {
    ...
}

In JavaScript, there are no named arguments like in PHP. However, you can pass an object instead, such as:

load_img(src, alt, '#zoom-cover', {other_src: 'something'});

This approach is commonly seen in jQuery Plugins.

What is the appropriate method to call a function when a parameter is unnecessary?

Answer №1

Answer №2

Answer №3

Answer №4

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