Timeline for Javascript - Difference between function declarations
Current License: CC BY-SA 3.0
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13 events
| when toggle format | what | by | license | comment | |
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| Aug 13, 2017 at 13:01 | history | edited | Cœur | CC BY-SA 3.0 |
spelling in title
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| Apr 13, 2017 at 9:22 | vote | accept | hjalpmig | ||
| Apr 13, 2017 at 9:18 | review | Close votes | |||
| Apr 17, 2017 at 0:04 | |||||
| Apr 13, 2017 at 9:14 | answer | added | Muhammad Qasim | timeline score: 0 | |
| Apr 13, 2017 at 9:13 | comment | added | Arg0n |
@Quentin Still not impossible to add the call inside. Code is usually modifiable. He does not need to put it outsite fun1 to be able to call it as you state. With your logic fun1 is impossible to call, since he does not call it with the shown code.
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| Apr 13, 2017 at 9:12 | comment | added | Saurabh Tiwari | What you do in the first case is an example of closure. Read about JavaScript closures and you should be clear about it | |
| Apr 13, 2017 at 9:08 | answer | added | m1kael | timeline score: 0 | |
| Apr 13, 2017 at 9:07 | comment | added | Quentin |
@Arg0n — It isn't called from inside fun1
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| Apr 13, 2017 at 9:04 | answer | added | MysterX | timeline score: 2 | |
| Apr 13, 2017 at 9:03 | comment | added | Arg0n |
@Quentin Impossible from outside fun1, inside it's not.
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| Apr 13, 2017 at 9:03 | comment | added | user3004449 | Second example is better for re-usability. Now you can call function fun2 again in other functions. | |
| Apr 13, 2017 at 9:02 | comment | added | Quentin |
Well ... it is impossible to call fun2 in the first example.
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| Apr 13, 2017 at 9:01 | history | asked | hjalpmig | CC BY-SA 3.0 |