Why does JavaScript ES6 function calls would not work as expected

The following snipped is given:

function output() {

  return "<p>normal function</p>";  

}



//first version

document.write(output());

//second version

document.write(() => {return "<p>arrow-function</p>"});

I was wondering, why a normal function would print correctly in the example above, but the second approach would just print the translated function declaration like that:

"function () {return "



arrow-function

";} "

Can you explain me that?

asked Nov 13 '18 at 22:12

Marc M

5281623

5

You are writing the method, and not the result of the method, like you do with output(). To make it work, you'll need to invoke the arrow function document.write((() => {return "arrow-function"})()) would work.

– Ori Drori
Nov 13 '18 at 22:14

2

Anyway: I've just came back here and see all the answers and comments now which must be happened within the last few minutes. So you really should calm down since I've not voted anything yet @jfadich

– Marc M
Nov 13 '18 at 22:36

There are a lot of helpful and correct answers below - I had to select one so I chose the one which has the best quality in formatting and has the most extensive explanation.

– Marc M
Nov 13 '18 at 22:46

add a comment |

The following snipped is given:

function output() {

  return "<p>normal function</p>";  

}



//first version

document.write(output());

//second version

document.write(() => {return "<p>arrow-function</p>"});

I was wondering, why a normal function would print correctly in the example above, but the second approach would just print the translated function declaration like that:

"function () {return "



arrow-function

";} "

Can you explain me that?

asked Nov 13 '18 at 22:12

Marc M

5281623

5

You are writing the method, and not the result of the method, like you do with output(). To make it work, you'll need to invoke the arrow function document.write((() => {return "arrow-function"})()) would work.

– Ori Drori
Nov 13 '18 at 22:14

2

Anyway: I've just came back here and see all the answers and comments now which must be happened within the last few minutes. So you really should calm down since I've not voted anything yet @jfadich

– Marc M
Nov 13 '18 at 22:36

There are a lot of helpful and correct answers below - I had to select one so I chose the one which has the best quality in formatting and has the most extensive explanation.

– Marc M
Nov 13 '18 at 22:46

add a comment |

The following snipped is given:

function output() {

  return "<p>normal function</p>";  

}



//first version

document.write(output());

//second version

document.write(() => {return "<p>arrow-function</p>"});

I was wondering, why a normal function would print correctly in the example above, but the second approach would just print the translated function declaration like that:

"function () {return "



arrow-function

";} "

Can you explain me that?

asked Nov 13 '18 at 22:12

Marc M

5281623

The following snipped is given:

function output() {

  return "<p>normal function</p>";  

}



//first version

document.write(output());

//second version

document.write(() => {return "<p>arrow-function</p>"});

I was wondering, why a normal function would print correctly in the example above, but the second approach would just print the translated function declaration like that:

"function () {return "



arrow-function

";} "

Can you explain me that?

javascript ecmascript-6

asked Nov 13 '18 at 22:12

Marc M

5281623

asked Nov 13 '18 at 22:12

Marc M

5281623

asked Nov 13 '18 at 22:12

Marc M

5281623

asked Nov 13 '18 at 22:12

Marc M

5281623

asked Nov 13 '18 at 22:12

Marc M

5281623

5

You are writing the method, and not the result of the method, like you do with output(). To make it work, you'll need to invoke the arrow function document.write((() => {return "arrow-function"})()) would work.

– Ori Drori
Nov 13 '18 at 22:14

2

Anyway: I've just came back here and see all the answers and comments now which must be happened within the last few minutes. So you really should calm down since I've not voted anything yet @jfadich

– Marc M
Nov 13 '18 at 22:36

There are a lot of helpful and correct answers below - I had to select one so I chose the one which has the best quality in formatting and has the most extensive explanation.

– Marc M
Nov 13 '18 at 22:46

add a comment |

5

You are writing the method, and not the result of the method, like you do with output(). To make it work, you'll need to invoke the arrow function document.write((() => {return "arrow-function"})()) would work.

– Ori Drori
Nov 13 '18 at 22:14

2

Anyway: I've just came back here and see all the answers and comments now which must be happened within the last few minutes. So you really should calm down since I've not voted anything yet @jfadich

– Marc M
Nov 13 '18 at 22:36

There are a lot of helpful and correct answers below - I had to select one so I chose the one which has the best quality in formatting and has the most extensive explanation.

– Marc M
Nov 13 '18 at 22:46

You are writing the method, and not the result of the method, like you do with output(). To make it work, you'll need to invoke the arrow function document.write((() => {return "arrow-function"})()) would work.

– Ori Drori
Nov 13 '18 at 22:14

Anyway: I've just came back here and see all the answers and comments now which must be happened within the last few minutes. So you really should calm down since I've not voted anything yet @jfadich

– Marc M
Nov 13 '18 at 22:36

There are a lot of helpful and correct answers below - I had to select one so I chose the one which has the best quality in formatting and has the most extensive explanation.

– Marc M
Nov 13 '18 at 22:46

add a comment |

4 Answers
4

active

oldest

votes

You aren't actually invoking the second function as you are with the first. If you wrap the function and invoke it you'll get the expected result.

function output() {

  return "<p>normal function</p>";  

}



//first version

document.write(output());

//second version

document.write(() => {return "<p>arrow-function</p>"});



//should be

document.write((() => {return "<p>arrow-function</p>"})());

answered Nov 13 '18 at 22:16

Jordan S

3,069921

add a comment |

It's because you are creating the function but not calling it.

To call the anonymous arrow function you should do:

(() => {return "<p>arrow-function</p>"})()

answered Nov 13 '18 at 22:15

Ariel Alvarado

682311

add a comment |

You are telling document.write to output a function definition that you are passing and you aren't really calling that arrow function so it isn't returning a value. You could do this instead and get the results you are expecting

var f = () => {return "<p>arrow-function</p>"};



document.write(f())

answered Nov 13 '18 at 22:16

Matti Price

2,002819

add a comment |

It is because the first parameter is a function call, whereas the second is only a function.

For the second call to work as the first, you would do:

document.write( (() => {return "<p>arrow-function</p>"})() );

This syntax above is pretty messy: you need the extra () to call the function. On top of that, you need to surround all the arrow function definition within parentheses due to language syntax requirements.

This way to immediately call functions from the very definition was really popular before ES2015 and is known as IIFE's. You usually see it with function statements but you can immediately call arrow functions too.

On the other hand, for the first line of your snippet to work as your original second, you would do instead:

document.write(output);

So now you're passing just two function definitions to document.write.

edited Nov 15 '18 at 15:50

answered Nov 13 '18 at 22:16

Sergeon

2,505818

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53290287%2fwhy-does-javascript-es6-function-calls-would-not-work-as-expected%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

You aren't actually invoking the second function as you are with the first. If you wrap the function and invoke it you'll get the expected result.

function output() {

  return "<p>normal function</p>";  

}



//first version

document.write(output());

//second version

document.write(() => {return "<p>arrow-function</p>"});



//should be

document.write((() => {return "<p>arrow-function</p>"})());

answered Nov 13 '18 at 22:16

Jordan S

3,069921

add a comment |

You aren't actually invoking the second function as you are with the first. If you wrap the function and invoke it you'll get the expected result.

function output() {

  return "<p>normal function</p>";  

}



//first version

document.write(output());

//second version

document.write(() => {return "<p>arrow-function</p>"});



//should be

document.write((() => {return "<p>arrow-function</p>"})());

answered Nov 13 '18 at 22:16

Jordan S

3,069921

add a comment |

You aren't actually invoking the second function as you are with the first. If you wrap the function and invoke it you'll get the expected result.

function output() {

  return "<p>normal function</p>";  

}



//first version

document.write(output());

//second version

document.write(() => {return "<p>arrow-function</p>"});



//should be

document.write((() => {return "<p>arrow-function</p>"})());

answered Nov 13 '18 at 22:16

Jordan S

3,069921

You aren't actually invoking the second function as you are with the first. If you wrap the function and invoke it you'll get the expected result.

function output() {

  return "<p>normal function</p>";  

}



//first version

document.write(output());

//second version

document.write(() => {return "<p>arrow-function</p>"});



//should be

document.write((() => {return "<p>arrow-function</p>"})());

function output() {

  return "<p>normal function</p>";  

}



//first version

document.write(output());

//second version

document.write(() => {return "<p>arrow-function</p>"});



//should be

document.write((() => {return "<p>arrow-function</p>"})());

function output() {

  return "<p>normal function</p>";  

}



//first version

document.write(output());

//second version

document.write(() => {return "<p>arrow-function</p>"});



//should be

document.write((() => {return "<p>arrow-function</p>"})());

answered Nov 13 '18 at 22:16

Jordan S

3,069921

answered Nov 13 '18 at 22:16

Jordan S

3,069921

answered Nov 13 '18 at 22:16

Jordan S

3,069921

answered Nov 13 '18 at 22:16

Jordan S

3,069921

add a comment |

It's because you are creating the function but not calling it.

To call the anonymous arrow function you should do:

(() => {return "<p>arrow-function</p>"})()

answered Nov 13 '18 at 22:15

Ariel Alvarado

682311

add a comment |

It's because you are creating the function but not calling it.

To call the anonymous arrow function you should do:

(() => {return "<p>arrow-function</p>"})()

answered Nov 13 '18 at 22:15

Ariel Alvarado

682311

add a comment |

It's because you are creating the function but not calling it.

To call the anonymous arrow function you should do:

(() => {return "<p>arrow-function</p>"})()

answered Nov 13 '18 at 22:15

Ariel Alvarado

682311

It's because you are creating the function but not calling it.

To call the anonymous arrow function you should do:

(() => {return "<p>arrow-function</p>"})()

answered Nov 13 '18 at 22:15

Ariel Alvarado

682311

answered Nov 13 '18 at 22:15

Ariel Alvarado

682311

answered Nov 13 '18 at 22:15

Ariel Alvarado

682311

answered Nov 13 '18 at 22:15

Ariel Alvarado

682311

add a comment |

var f = () => {return "<p>arrow-function</p>"};



document.write(f())

answered Nov 13 '18 at 22:16

Matti Price

2,002819

add a comment |

var f = () => {return "<p>arrow-function</p>"};



document.write(f())

answered Nov 13 '18 at 22:16

Matti Price

2,002819

add a comment |

var f = () => {return "<p>arrow-function</p>"};



document.write(f())

answered Nov 13 '18 at 22:16

Matti Price

2,002819

var f = () => {return "<p>arrow-function</p>"};



document.write(f())

answered Nov 13 '18 at 22:16

Matti Price

2,002819

answered Nov 13 '18 at 22:16

Matti Price

2,002819

answered Nov 13 '18 at 22:16

Matti Price

2,002819

answered Nov 13 '18 at 22:16

Matti Price

2,002819

add a comment |

It is because the first parameter is a function call, whereas the second is only a function.

For the second call to work as the first, you would do:

document.write( (() => {return "<p>arrow-function</p>"})() );

On the other hand, for the first line of your snippet to work as your original second, you would do instead:

document.write(output);

So now you're passing just two function definitions to document.write.

edited Nov 15 '18 at 15:50

answered Nov 13 '18 at 22:16

Sergeon

2,505818

add a comment |

It is because the first parameter is a function call, whereas the second is only a function.

For the second call to work as the first, you would do:

document.write( (() => {return "<p>arrow-function</p>"})() );

On the other hand, for the first line of your snippet to work as your original second, you would do instead:

document.write(output);

So now you're passing just two function definitions to document.write.

edited Nov 15 '18 at 15:50

answered Nov 13 '18 at 22:16

Sergeon

2,505818

add a comment |

It is because the first parameter is a function call, whereas the second is only a function.

For the second call to work as the first, you would do:

document.write( (() => {return "<p>arrow-function</p>"})() );

On the other hand, for the first line of your snippet to work as your original second, you would do instead:

document.write(output);

So now you're passing just two function definitions to document.write.

edited Nov 15 '18 at 15:50

answered Nov 13 '18 at 22:16

Sergeon

2,505818

It is because the first parameter is a function call, whereas the second is only a function.

For the second call to work as the first, you would do:

document.write( (() => {return "<p>arrow-function</p>"})() );

On the other hand, for the first line of your snippet to work as your original second, you would do instead:

document.write(output);

So now you're passing just two function definitions to document.write.

edited Nov 15 '18 at 15:50

answered Nov 13 '18 at 22:16

Sergeon

2,505818

edited Nov 15 '18 at 15:50

answered Nov 13 '18 at 22:16

Sergeon

2,505818

answered Nov 13 '18 at 22:16

Sergeon

2,505818

answered Nov 13 '18 at 22:16

Sergeon

2,505818

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Vfrdtyky