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Xcode 10.1 “return” statement doesn't stop function execution [duplicate]

early return from a void-func in Swift?Xcode 4 Error: Error Starting ExecutableXcode doesn't show code when stops at breakpointsUIScrollView Scrollable Content Size AmbiguitySwift Beta performance: sorting arraysSwift Parse - Method doesn't stop executingThe amount of Xcode tasks increase after every buildAdding localizable strings on xcode 10.1lldb-rpc-server in Xcode Version is 10.1?What's the solution for “error: Couldn't IRGen expression, no additional error” on Xcode 10.1?cloudkit disabled in capabilities after upgrade to xcode 10.1

1

This question already has an answer here:

• 
  early return from a void-func in Swift?
  
  5 answers
  
  

The return statement in Xcode 10.1 is not honoured by debugger,

For eg.,

 func doSomething() 

 print("Task A")
 return

 print("Task B")
 

This prints

Task A
Task B //This is not expected to be printed as we have a `return` before this line 

Can someone help me!

ios swift xcode xcode10.1

share|improve this question

asked yesterday

SaifSaif

1,2831923

marked as duplicate by Martin R ios
Users with the  ios badge can single-handedly close ios questions as duplicates and reopen them as needed.

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yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That should not happen. Try a clean and rebuild. Sometimes Xcode loses its mind and does not rebuild your source files after you make changes.
  
  – Duncan C
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @DuncanC: It does happen (see the linked-to Q&A). Swift executes the statement return print("Task B"), which prints the string and returns void. – Whether that should be considered a bug or not is a different story. Inserting a semicolon helps.
  
  – Martin R
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MartinR, well it's definitely not a bug, just a natural quirk of non-mandatory-semicolon syntax.
  
  – user28434
  yesterday
  

  
  
  
  

add a comment | 

1

This question already has an answer here:

• 
  early return from a void-func in Swift?
  
  5 answers
  
  

The return statement in Xcode 10.1 is not honoured by debugger,

For eg.,

 func doSomething() 

 print("Task A")
 return

 print("Task B")
 

This prints

Task A
Task B //This is not expected to be printed as we have a `return` before this line 

Can someone help me!

ios swift xcode xcode10.1

share|improve this question

asked yesterday

SaifSaif

1,2831923

marked as duplicate by Martin R ios
Users with the  ios badge can single-handedly close ios questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
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$hover.hover(
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);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That should not happen. Try a clean and rebuild. Sometimes Xcode loses its mind and does not rebuild your source files after you make changes.
  
  – Duncan C
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @DuncanC: It does happen (see the linked-to Q&A). Swift executes the statement return print("Task B"), which prints the string and returns void. – Whether that should be considered a bug or not is a different story. Inserting a semicolon helps.
  
  – Martin R
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MartinR, well it's definitely not a bug, just a natural quirk of non-mandatory-semicolon syntax.
  
  – user28434
  yesterday
  

  
  
  
  

add a comment | 

This question already has an answer here:

• 
  early return from a void-func in Swift?
  
  5 answers
  
  

The return statement in Xcode 10.1 is not honoured by debugger,

For eg.,

 func doSomething() 

 print("Task A")
 return

 print("Task B")
 

This prints

Task A
Task B //This is not expected to be printed as we have a `return` before this line 

Can someone help me!

ios swift xcode xcode10.1

share|improve this question

asked yesterday

SaifSaif

1,2831923

 func doSomething() 

 print("Task A")
 return

 print("Task B")
 

Task A
Task B //This is not expected to be printed as we have a `return` before this line 

share|improve this question

asked yesterday

SaifSaif

1,2831923

share|improve this question

asked yesterday

SaifSaif

1,2831923

asked yesterday

SaifSaif

1,2831923

asked yesterday

SaifSaif

1,2831923

2 Answers
2

active

oldest

votes

6

Because expression after return is treated as argument of return.

So your code understood by compiler as:

func doSomething() 
 print("Task A")
 return print("Task B")

To prevent it you can use semicolon to explicitly separate this expressions.

Like that:

func doSomething() 
 print("Task A")
 return;
 print("Task B")

share|improve this answer

answered yesterday

ManWithBearManWithBear

684519

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  ... as explained in stackoverflow.com/q/48887696/1187415 :)
  
  – Martin R
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MartinR yeah, I failed to find this question :D
  
  – ManWithBear
  yesterday
  

  
  
  
  

add a comment | 

-4

try this :

func doSomething() 


 return print("Task A")

 print("Task B")

the function itself does not wait for a value to get return, thats why in your case return wont stop the execution of the function,

but if your implement the return the way i showed above, it will return the print("task a") and will stop right away

share|improve this answer

edited yesterday

answered yesterday

Sanad BarjawiSanad Barjawi

10711

add a comment | 

6

Because expression after return is treated as argument of return.

So your code understood by compiler as:

func doSomething() 
 print("Task A")
 return print("Task B")

To prevent it you can use semicolon to explicitly separate this expressions.

Like that:

func doSomething() 
 print("Task A")
 return;
 print("Task B")

share|improve this answer

answered yesterday

ManWithBearManWithBear

684519

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  ... as explained in stackoverflow.com/q/48887696/1187415 :)
  
  – Martin R
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MartinR yeah, I failed to find this question :D
  
  – ManWithBear
  yesterday
  

  
  
  
  

add a comment | 

6

Because expression after return is treated as argument of return.

So your code understood by compiler as:

func doSomething() 
 print("Task A")
 return print("Task B")

To prevent it you can use semicolon to explicitly separate this expressions.

Like that:

func doSomething() 
 print("Task A")
 return;
 print("Task B")

share|improve this answer

answered yesterday

ManWithBearManWithBear

684519

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  ... as explained in stackoverflow.com/q/48887696/1187415 :)
  
  – Martin R
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MartinR yeah, I failed to find this question :D
  
  – ManWithBear
  yesterday
  

  
  
  
  

add a comment | 

Because expression after return is treated as argument of return.

So your code understood by compiler as:

func doSomething() 
 print("Task A")
 return print("Task B")

To prevent it you can use semicolon to explicitly separate this expressions.

Like that:

func doSomething() 
 print("Task A")
 return;
 print("Task B")

share|improve this answer

answered yesterday

ManWithBearManWithBear

684519

func doSomething() 
 print("Task A")
 return print("Task B")

func doSomething() 
 print("Task A")
 return;
 print("Task B")

share|improve this answer

answered yesterday

ManWithBearManWithBear

684519

answered yesterday

ManWithBearManWithBear

684519

answered yesterday

ManWithBearManWithBear

684519

-4

try this :

func doSomething() 


 return print("Task A")

 print("Task B")

the function itself does not wait for a value to get return, thats why in your case return wont stop the execution of the function,

but if your implement the return the way i showed above, it will return the print("task a") and will stop right away

share|improve this answer

edited yesterday

answered yesterday

Sanad BarjawiSanad Barjawi

10711

add a comment | 

-4

try this :

func doSomething() 


 return print("Task A")

 print("Task B")

the function itself does not wait for a value to get return, thats why in your case return wont stop the execution of the function,

but if your implement the return the way i showed above, it will return the print("task a") and will stop right away

share|improve this answer

edited yesterday

answered yesterday

Sanad BarjawiSanad Barjawi

10711

add a comment | 

try this :

func doSomething() 


 return print("Task A")

 print("Task B")

the function itself does not wait for a value to get return, thats why in your case return wont stop the execution of the function,

but if your implement the return the way i showed above, it will return the print("task a") and will stop right away

share|improve this answer

edited yesterday

answered yesterday

Sanad BarjawiSanad Barjawi

10711

func doSomething() 


 return print("Task A")

 print("Task B")

share|improve this answer

edited yesterday

answered yesterday

Sanad BarjawiSanad Barjawi

10711

answered yesterday

Sanad BarjawiSanad Barjawi

10711

answered yesterday

Sanad BarjawiSanad Barjawi

10711