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how a variable declaration works inside a case in switch statement [duplicate]

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2

This question already has an answer here:

• 
  Declaring and initializing variables within Java switches
  
  5 answers
  
  

I'm having trouble understanding how the switch statement works. The following code works even the value of a is 4. I understand that each case inside a switch statement is not a block with its own local scope but does that mean that the variable x was declared even a is not equal to 2?

int a = 2;
switch(a)
 case 2:
 int x = 4;
 case 3:
 x = 5;
 case 4: 
 x = 7; 

java switch-statement

share|improve this question

edited Mar 23 at 2:16

Tomin B Azhakathu

2,09011022

asked Mar 23 at 1:30

glennmarkglennmark

708

marked as duplicate by Community♦ Mar 23 at 2:11

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.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  As noted in the answer to the other question it boils down to this: A declaration isn't really "executable" although initialization is. So even if int x = 4 is never executed, you've still declared a variable x in the scope of the switch. So x will be declared even if case 2 never runs
  
  – Nikos Tzianas
  Mar 23 at 2:07
  
  

  
  
  
  

add a comment | 

2

This question already has an answer here:

• 
  Declaring and initializing variables within Java switches
  
  5 answers
  
  

I'm having trouble understanding how the switch statement works. The following code works even the value of a is 4. I understand that each case inside a switch statement is not a block with its own local scope but does that mean that the variable x was declared even a is not equal to 2?

int a = 2;
switch(a)
 case 2:
 int x = 4;
 case 3:
 x = 5;
 case 4: 
 x = 7; 

java switch-statement

share|improve this question

edited Mar 23 at 2:16

Tomin B Azhakathu

2,09011022

asked Mar 23 at 1:30

glennmarkglennmark

708

marked as duplicate by Community♦ Mar 23 at 2:11

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.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  As noted in the answer to the other question it boils down to this: A declaration isn't really "executable" although initialization is. So even if int x = 4 is never executed, you've still declared a variable x in the scope of the switch. So x will be declared even if case 2 never runs
  
  – Nikos Tzianas
  Mar 23 at 2:07
  
  

  
  
  
  

add a comment | 

This question already has an answer here:

• 
  Declaring and initializing variables within Java switches
  
  5 answers
  
  

I'm having trouble understanding how the switch statement works. The following code works even the value of a is 4. I understand that each case inside a switch statement is not a block with its own local scope but does that mean that the variable x was declared even a is not equal to 2?

int a = 2;
switch(a)
 case 2:
 int x = 4;
 case 3:
 x = 5;
 case 4: 
 x = 7; 

java switch-statement

share|improve this question

edited Mar 23 at 2:16

Tomin B Azhakathu

2,09011022

asked Mar 23 at 1:30

glennmarkglennmark

708

int a = 2;
switch(a)
 case 2:
 int x = 4;
 case 3:
 x = 5;
 case 4: 
 x = 7; 

share|improve this question

edited Mar 23 at 2:16

Tomin B Azhakathu

2,09011022

asked Mar 23 at 1:30

glennmarkglennmark

708

share|improve this question

edited Mar 23 at 2:16

Tomin B Azhakathu

2,09011022

asked Mar 23 at 1:30

glennmarkglennmark

708

edited Mar 23 at 2:16

Tomin B Azhakathu

2,09011022

edited Mar 23 at 2:16

Tomin B Azhakathu

2,09011022

asked Mar 23 at 1:30

glennmarkglennmark

708

asked Mar 23 at 1:30

glennmarkglennmark

708

1 Answer
1

active

oldest

votes

0

Case 1

 int a = 2;
 switch(a)

 case 2:
 int x = 4;
 // executes
 case 3:
 x = 5;
 // executes
 case 4: 
 x = 7;
 // executes

 

in this value of x will be 7 because the switch won't break if it find a matching case.

Case 2

If you want to stop execution on 2 do the following

 int a = 2;
 switch(a)

 case 2:
 int x = 4; // variable declaration works even if case wont match. Validity inside switch for every line of code after this line
 break;
 // executes and break
 case 3:
 x = 5;
 break;
 // executes and break
 case 4: 
 x = 7;
 break;
 // executes and break
 default:
 // write default case here
 

You need not use int in every x because of the scope of the declaration throughout the scope.

The scope of the declaration of a variable in the switch will affect the following statements or line of code in the switch. It's because, if the variable declaration it will if the scope is not defined by braces.

Case 3

 int a = 3;
 switch(a)

 case 2:
 
 int x = 4;
 break; // u can skip this break to. Then also it will throw error.
 // x is local variable inside this braces
 
 // executes
 case 3:
 x = 5;
 // error
 case 4: 
 x = 7;
 // error
 System.out.println(""+x); 


 

In the above case, it will through the error. Because its scope is only in that case.
In this scenario, the scope of a variable is declared inside the braces. so outside that braces, x won't work. Because x is a local variable in that case condition.

share|improve this answer

edited Mar 23 at 2:02

answered Mar 23 at 1:38

Tomin B AzhakathuTomin B Azhakathu

2,09011022

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  i think you misunderstood the question.. if a == 4, then x = 7.. but there must be some error because x is only declared in case 2.. my question is why does it not throw an error..
  
  – glennmark
  Mar 23 at 1:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The scope of decleartion of a variable in switch will affect the following statements or line of code in switch
  
  – Tomin B Azhakathu
  Mar 23 at 1:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @glennmark this wont throw error.
  
  – Tomin B Azhakathu
  Mar 23 at 1:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  do you mean that every variable declaration inside a switch statement is invoked even though they don't fall in the right case?
  
  – glennmark
  Mar 23 at 1:49
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @glennmark Can you check the explaination with comment is enough. Else we can discuss in chat
  
  – Tomin B Azhakathu
  Mar 23 at 2:03
  

  
  
  
  

 | 
show 7 more comments

0

Case 1

 int a = 2;
 switch(a)

 case 2:
 int x = 4;
 // executes
 case 3:
 x = 5;
 // executes
 case 4: 
 x = 7;
 // executes

 

in this value of x will be 7 because the switch won't break if it find a matching case.

Case 2

If you want to stop execution on 2 do the following

 int a = 2;
 switch(a)

 case 2:
 int x = 4; // variable declaration works even if case wont match. Validity inside switch for every line of code after this line
 break;
 // executes and break
 case 3:
 x = 5;
 break;
 // executes and break
 case 4: 
 x = 7;
 break;
 // executes and break
 default:
 // write default case here
 

You need not use int in every x because of the scope of the declaration throughout the scope.

The scope of the declaration of a variable in the switch will affect the following statements or line of code in the switch. It's because, if the variable declaration it will if the scope is not defined by braces.

Case 3

 int a = 3;
 switch(a)

 case 2:
 
 int x = 4;
 break; // u can skip this break to. Then also it will throw error.
 // x is local variable inside this braces
 
 // executes
 case 3:
 x = 5;
 // error
 case 4: 
 x = 7;
 // error
 System.out.println(""+x); 


 

In the above case, it will through the error. Because its scope is only in that case.
In this scenario, the scope of a variable is declared inside the braces. so outside that braces, x won't work. Because x is a local variable in that case condition.

share|improve this answer

edited Mar 23 at 2:02

answered Mar 23 at 1:38

Tomin B AzhakathuTomin B Azhakathu

2,09011022

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  i think you misunderstood the question.. if a == 4, then x = 7.. but there must be some error because x is only declared in case 2.. my question is why does it not throw an error..
  
  – glennmark
  Mar 23 at 1:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The scope of decleartion of a variable in switch will affect the following statements or line of code in switch
  
  – Tomin B Azhakathu
  Mar 23 at 1:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @glennmark this wont throw error.
  
  – Tomin B Azhakathu
  Mar 23 at 1:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  do you mean that every variable declaration inside a switch statement is invoked even though they don't fall in the right case?
  
  – glennmark
  Mar 23 at 1:49
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @glennmark Can you check the explaination with comment is enough. Else we can discuss in chat
  
  – Tomin B Azhakathu
  Mar 23 at 2:03
  

  
  
  
  

 | 
show 7 more comments

0

Case 1

 int a = 2;
 switch(a)

 case 2:
 int x = 4;
 // executes
 case 3:
 x = 5;
 // executes
 case 4: 
 x = 7;
 // executes

 

in this value of x will be 7 because the switch won't break if it find a matching case.

Case 2

If you want to stop execution on 2 do the following

 int a = 2;
 switch(a)

 case 2:
 int x = 4; // variable declaration works even if case wont match. Validity inside switch for every line of code after this line
 break;
 // executes and break
 case 3:
 x = 5;
 break;
 // executes and break
 case 4: 
 x = 7;
 break;
 // executes and break
 default:
 // write default case here
 

You need not use int in every x because of the scope of the declaration throughout the scope.

The scope of the declaration of a variable in the switch will affect the following statements or line of code in the switch. It's because, if the variable declaration it will if the scope is not defined by braces.

Case 3

 int a = 3;
 switch(a)

 case 2:
 
 int x = 4;
 break; // u can skip this break to. Then also it will throw error.
 // x is local variable inside this braces
 
 // executes
 case 3:
 x = 5;
 // error
 case 4: 
 x = 7;
 // error
 System.out.println(""+x); 


 

In the above case, it will through the error. Because its scope is only in that case.
In this scenario, the scope of a variable is declared inside the braces. so outside that braces, x won't work. Because x is a local variable in that case condition.

share|improve this answer

edited Mar 23 at 2:02

answered Mar 23 at 1:38

Tomin B AzhakathuTomin B Azhakathu

2,09011022

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  i think you misunderstood the question.. if a == 4, then x = 7.. but there must be some error because x is only declared in case 2.. my question is why does it not throw an error..
  
  – glennmark
  Mar 23 at 1:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The scope of decleartion of a variable in switch will affect the following statements or line of code in switch
  
  – Tomin B Azhakathu
  Mar 23 at 1:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @glennmark this wont throw error.
  
  – Tomin B Azhakathu
  Mar 23 at 1:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  do you mean that every variable declaration inside a switch statement is invoked even though they don't fall in the right case?
  
  – glennmark
  Mar 23 at 1:49
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @glennmark Can you check the explaination with comment is enough. Else we can discuss in chat
  
  – Tomin B Azhakathu
  Mar 23 at 2:03
  

  
  
  
  

 | 
show 7 more comments

Case 1

 int a = 2;
 switch(a)

 case 2:
 int x = 4;
 // executes
 case 3:
 x = 5;
 // executes
 case 4: 
 x = 7;
 // executes

 

in this value of x will be 7 because the switch won't break if it find a matching case.

Case 2

If you want to stop execution on 2 do the following

 int a = 2;
 switch(a)

 case 2:
 int x = 4; // variable declaration works even if case wont match. Validity inside switch for every line of code after this line
 break;
 // executes and break
 case 3:
 x = 5;
 break;
 // executes and break
 case 4: 
 x = 7;
 break;
 // executes and break
 default:
 // write default case here
 

You need not use int in every x because of the scope of the declaration throughout the scope.

The scope of the declaration of a variable in the switch will affect the following statements or line of code in the switch. It's because, if the variable declaration it will if the scope is not defined by braces.

Case 3

 int a = 3;
 switch(a)

 case 2:
 
 int x = 4;
 break; // u can skip this break to. Then also it will throw error.
 // x is local variable inside this braces
 
 // executes
 case 3:
 x = 5;
 // error
 case 4: 
 x = 7;
 // error
 System.out.println(""+x); 


 

In the above case, it will through the error. Because its scope is only in that case.
In this scenario, the scope of a variable is declared inside the braces. so outside that braces, x won't work. Because x is a local variable in that case condition.

share|improve this answer

edited Mar 23 at 2:02

answered Mar 23 at 1:38

Tomin B AzhakathuTomin B Azhakathu

2,09011022

 int a = 2;
 switch(a)

 case 2:
 int x = 4;
 // executes
 case 3:
 x = 5;
 // executes
 case 4: 
 x = 7;
 // executes

 

 int a = 2;
 switch(a)

 case 2:
 int x = 4; // variable declaration works even if case wont match. Validity inside switch for every line of code after this line
 break;
 // executes and break
 case 3:
 x = 5;
 break;
 // executes and break
 case 4: 
 x = 7;
 break;
 // executes and break
 default:
 // write default case here
 

 int a = 3;
 switch(a)

 case 2:
 
 int x = 4;
 break; // u can skip this break to. Then also it will throw error.
 // x is local variable inside this braces
 
 // executes
 case 3:
 x = 5;
 // error
 case 4: 
 x = 7;
 // error
 System.out.println(""+x); 


 

share|improve this answer

edited Mar 23 at 2:02

answered Mar 23 at 1:38

Tomin B AzhakathuTomin B Azhakathu

2,09011022

answered Mar 23 at 1:38

Tomin B AzhakathuTomin B Azhakathu

2,09011022

answered Mar 23 at 1:38

Tomin B AzhakathuTomin B Azhakathu

2,09011022