Conditional method chaining in java

What is the best possible way to chain methods together? In my scenario, there are four methods. If the output of the first method is true, it has to call the second method.

For Example:

flag = method1();

if (flag){

  flag = method2();

}

if (flag){

  method3();

} 

// and so on for next methods ..

edited Dec 21 '18 at 20:00

Federico Peralta Schaffner

22.3k43573

asked Dec 21 '18 at 19:42

user10821509

624

Use a nested set of conditional operators. It won't be legible but it will work.

– nicomp
Dec 21 '18 at 19:46

Do you need to return flag after the last method has been executed?

– Federico Peralta Schaffner
Dec 21 '18 at 19:50

1

"best" by what metric?

– TylerH
Dec 21 '18 at 22:52

add a comment |

What is the best possible way to chain methods together? In my scenario, there are four methods. If the output of the first method is true, it has to call the second method.

For Example:

flag = method1();

if (flag){

  flag = method2();

}

if (flag){

  method3();

} 

// and so on for next methods ..

edited Dec 21 '18 at 20:00

Federico Peralta Schaffner

22.3k43573

asked Dec 21 '18 at 19:42

user10821509

624

Use a nested set of conditional operators. It won't be legible but it will work.

– nicomp
Dec 21 '18 at 19:46

Do you need to return flag after the last method has been executed?

– Federico Peralta Schaffner
Dec 21 '18 at 19:50

1

"best" by what metric?

– TylerH
Dec 21 '18 at 22:52

add a comment |

What is the best possible way to chain methods together? In my scenario, there are four methods. If the output of the first method is true, it has to call the second method.

For Example:

flag = method1();

if (flag){

  flag = method2();

}

if (flag){

  method3();

} 

// and so on for next methods ..

edited Dec 21 '18 at 20:00

Federico Peralta Schaffner

22.3k43573

asked Dec 21 '18 at 19:42

user10821509

624

What is the best possible way to chain methods together? In my scenario, there are four methods. If the output of the first method is true, it has to call the second method.

For Example:

flag = method1();

if (flag){

  flag = method2();

}

if (flag){

  method3();

} 

// and so on for next methods ..

java

edited Dec 21 '18 at 20:00

Federico Peralta Schaffner

22.3k43573

asked Dec 21 '18 at 19:42

user10821509

624

edited Dec 21 '18 at 20:00

Federico Peralta Schaffner

22.3k43573

asked Dec 21 '18 at 19:42

user10821509

624

edited Dec 21 '18 at 20:00

Federico Peralta Schaffner

22.3k43573

edited Dec 21 '18 at 20:00

Federico Peralta Schaffner

22.3k43573

edited Dec 21 '18 at 20:00

Federico Peralta Schaffner

22.3k43573

asked Dec 21 '18 at 19:42

user10821509

624

asked Dec 21 '18 at 19:42

user10821509

624

asked Dec 21 '18 at 19:42

user10821509

624

Use a nested set of conditional operators. It won't be legible but it will work.

– nicomp
Dec 21 '18 at 19:46

Do you need to return flag after the last method has been executed?

– Federico Peralta Schaffner
Dec 21 '18 at 19:50

1

"best" by what metric?

– TylerH
Dec 21 '18 at 22:52

add a comment |

Use a nested set of conditional operators. It won't be legible but it will work.

– nicomp
Dec 21 '18 at 19:46

Do you need to return flag after the last method has been executed?

– Federico Peralta Schaffner
Dec 21 '18 at 19:50

1

"best" by what metric?

– TylerH
Dec 21 '18 at 22:52

Use a nested set of conditional operators. It won't be legible but it will work.

– nicomp
Dec 21 '18 at 19:46

Do you need to return flag after the last method has been executed?

– Federico Peralta Schaffner
Dec 21 '18 at 19:50

"best" by what metric?

– TylerH
Dec 21 '18 at 22:52

add a comment |

7 Answers
7

active

oldest

votes

Use the && logical AND operator:

if (method1() && method2() && method3() && method4()) {

    ........

}

Java evaluates this condition from left to right.

If any of the methods returns false, then the evaluation stops, because the final result is false (Short Circuit Evaluation).

edited Dec 21 '18 at 20:02

answered Dec 21 '18 at 19:46

forpas

10.4k1421

That doesn't put anything in flag.

– nicomp
Dec 21 '18 at 19:47

8

The flag is not needed this way.

– forpas
Dec 21 '18 at 19:48

This is hacker way :) But makes code, probably, hard to read.

– Koziołek
Dec 21 '18 at 19:54

6

@Koziołek If you know what you're doing, it is not hard to read.

– forpas
Dec 21 '18 at 19:55

2

@Koziołek as I said If you know what you're doing....

– forpas
Dec 21 '18 at 20:01

|
show 7 more comments

if flag is not needed somewhere later on then @forpas's answer is the way to go otherwise I'd do:

flag = method1() && method2() && method3() && method4()

edited Dec 21 '18 at 20:00

answered Dec 21 '18 at 19:54

Aomine

41.9k74071

add a comment |

To use the short circuit but keep it more readable, you can do

boolean flag = method1();

flag = flag && method2();

flag = flag && method3();

And so on. Note you can't use flag &= methodX() because that doesn't short circuit.

edited Dec 21 '18 at 21:30

answered Dec 21 '18 at 20:05

daniu

7,42521635

add a comment |

First and foremost I'd like to state that this answer is by no means more efficient than this answer or this answer rather it's just another way in which you can accomplish the task at hand while maintaining the short-shircuting requirement of yours.

So first step is to create a method as such:

public static boolean apply(BooleanSupplier... funcs){

      return Arrays.stream(funcs).allMatch(BooleanSupplier::getAsBoolean);

}

This function consumes any number of functions that have the signature () -> boolean i.e. a supplier (a function taking no inputs and returns a boolean result).

Then you can call it as follows using a method reference:

if(apply(Main::method1, 

         Main::method2, 

         Main::method3, 

         Main::method4)){ ... };

Where Main is the class where the methods are defined.

or lambda:

if(apply(() -> method1(), 

         () -> method2(), 

         () -> method3(), 

         () -> method4())){ ... };

edited Dec 21 '18 at 22:44

answered Dec 21 '18 at 22:27

Aomine

41.9k74071

This one is likely more readable if the list is more than 3 or 4.

– jpmc26
Dec 22 '18 at 1:18

add a comment |

Nested conditional operators would solve this:

flag = method1() ? (method2() ? method3(): false): false;

answered Dec 21 '18 at 19:50

nicomp

2,60621231

nice +1, you can simplify it to flag = method1() && (method2() && method3());. assuming flag is needed after the condtions then this is a good approach.

– Aomine
Dec 21 '18 at 19:51

@Aomine I think that would be better answer than this

– eis
Dec 21 '18 at 19:52

@Aomine but the conditional operator get so little respect! :)

– nicomp
Dec 21 '18 at 19:53

1

@nicomp right, i'll post it as an answer then :). #eis sure thing.

– Aomine
Dec 21 '18 at 19:54

add a comment |

You can use the && operator :

boolean flag = method1() && method2() && method3() && method4() && ...;

Do whatever with flag.

answered Dec 21 '18 at 20:03

fastcodejava

24.2k19109162

add a comment |

incase the methods truth values have to be saved or other side-effects have to happen:

boolean flag;

switch(1){

  case 1: flag = method1();  

          //additional code goes here



          if(!flag) break; 

  case 2: flag = method2();

          //additional code goes here



          if(!flag) break; 

     ....

  default: flag = method10();

          if(!flag) break; 

}

you shouldn't want this and it's... unexpected.
but this is a readable way of adding additional statements to each code path.

otherwise: do it differently or fall back on ifs.

answered Dec 21 '18 at 21:13

RecurringError

623

add a comment |

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7 Answers
7

active

oldest

votes

7 Answers
7

active

oldest

votes

Use the && logical AND operator:

if (method1() && method2() && method3() && method4()) {

    ........

}

Java evaluates this condition from left to right.

If any of the methods returns false, then the evaluation stops, because the final result is false (Short Circuit Evaluation).

edited Dec 21 '18 at 20:02

answered Dec 21 '18 at 19:46

forpas

10.4k1421

That doesn't put anything in flag.

– nicomp
Dec 21 '18 at 19:47

8

The flag is not needed this way.

– forpas
Dec 21 '18 at 19:48

This is hacker way :) But makes code, probably, hard to read.

– Koziołek
Dec 21 '18 at 19:54

6

@Koziołek If you know what you're doing, it is not hard to read.

– forpas
Dec 21 '18 at 19:55

2

@Koziołek as I said If you know what you're doing....

– forpas
Dec 21 '18 at 20:01

|
show 7 more comments

Use the && logical AND operator:

if (method1() && method2() && method3() && method4()) {

    ........

}

Java evaluates this condition from left to right.

If any of the methods returns false, then the evaluation stops, because the final result is false (Short Circuit Evaluation).

edited Dec 21 '18 at 20:02

answered Dec 21 '18 at 19:46

forpas

10.4k1421

That doesn't put anything in flag.

– nicomp
Dec 21 '18 at 19:47

8

The flag is not needed this way.

– forpas
Dec 21 '18 at 19:48

This is hacker way :) But makes code, probably, hard to read.

– Koziołek
Dec 21 '18 at 19:54

6

@Koziołek If you know what you're doing, it is not hard to read.

– forpas
Dec 21 '18 at 19:55

2

@Koziołek as I said If you know what you're doing....

– forpas
Dec 21 '18 at 20:01

|
show 7 more comments

Use the && logical AND operator:

if (method1() && method2() && method3() && method4()) {

    ........

}

Java evaluates this condition from left to right.

If any of the methods returns false, then the evaluation stops, because the final result is false (Short Circuit Evaluation).

edited Dec 21 '18 at 20:02

answered Dec 21 '18 at 19:46

forpas

10.4k1421

Use the && logical AND operator:

if (method1() && method2() && method3() && method4()) {

    ........

}

Java evaluates this condition from left to right.

If any of the methods returns false, then the evaluation stops, because the final result is false (Short Circuit Evaluation).

edited Dec 21 '18 at 20:02

answered Dec 21 '18 at 19:46

forpas

10.4k1421

edited Dec 21 '18 at 20:02

answered Dec 21 '18 at 19:46

forpas

10.4k1421

answered Dec 21 '18 at 19:46

forpas

10.4k1421

answered Dec 21 '18 at 19:46

forpas

10.4k1421

That doesn't put anything in flag.

– nicomp
Dec 21 '18 at 19:47

8

The flag is not needed this way.

– forpas
Dec 21 '18 at 19:48

This is hacker way :) But makes code, probably, hard to read.

– Koziołek
Dec 21 '18 at 19:54

6

@Koziołek If you know what you're doing, it is not hard to read.

– forpas
Dec 21 '18 at 19:55

2

@Koziołek as I said If you know what you're doing....

– forpas
Dec 21 '18 at 20:01

|
show 7 more comments

That doesn't put anything in flag.

– nicomp
Dec 21 '18 at 19:47

8

The flag is not needed this way.

– forpas
Dec 21 '18 at 19:48

This is hacker way :) But makes code, probably, hard to read.

– Koziołek
Dec 21 '18 at 19:54

6

@Koziołek If you know what you're doing, it is not hard to read.

– forpas
Dec 21 '18 at 19:55

2

@Koziołek as I said If you know what you're doing....

– forpas
Dec 21 '18 at 20:01

That doesn't put anything in flag.

– nicomp
Dec 21 '18 at 19:47

The flag is not needed this way.

– forpas
Dec 21 '18 at 19:48

This is hacker way :) But makes code, probably, hard to read.

– Koziołek
Dec 21 '18 at 19:54

@Koziołek If you know what you're doing, it is not hard to read.

– forpas
Dec 21 '18 at 19:55

@Koziołek as I said If you know what you're doing....

– forpas
Dec 21 '18 at 20:01

|
show 7 more comments

if flag is not needed somewhere later on then @forpas's answer is the way to go otherwise I'd do:

flag = method1() && method2() && method3() && method4()

edited Dec 21 '18 at 20:00

answered Dec 21 '18 at 19:54

Aomine

41.9k74071

add a comment |

if flag is not needed somewhere later on then @forpas's answer is the way to go otherwise I'd do:

flag = method1() && method2() && method3() && method4()

edited Dec 21 '18 at 20:00

answered Dec 21 '18 at 19:54

Aomine

41.9k74071

add a comment |

if flag is not needed somewhere later on then @forpas's answer is the way to go otherwise I'd do:

flag = method1() && method2() && method3() && method4()

edited Dec 21 '18 at 20:00

answered Dec 21 '18 at 19:54

Aomine

41.9k74071

if flag is not needed somewhere later on then @forpas's answer is the way to go otherwise I'd do:

flag = method1() && method2() && method3() && method4()

edited Dec 21 '18 at 20:00

answered Dec 21 '18 at 19:54

Aomine

41.9k74071

edited Dec 21 '18 at 20:00

answered Dec 21 '18 at 19:54

Aomine

41.9k74071

answered Dec 21 '18 at 19:54

Aomine

41.9k74071

answered Dec 21 '18 at 19:54

Aomine

41.9k74071

add a comment |

To use the short circuit but keep it more readable, you can do

boolean flag = method1();

flag = flag && method2();

flag = flag && method3();

And so on. Note you can't use flag &= methodX() because that doesn't short circuit.

edited Dec 21 '18 at 21:30

answered Dec 21 '18 at 20:05

daniu

7,42521635

add a comment |

To use the short circuit but keep it more readable, you can do

boolean flag = method1();

flag = flag && method2();

flag = flag && method3();

And so on. Note you can't use flag &= methodX() because that doesn't short circuit.

edited Dec 21 '18 at 21:30

answered Dec 21 '18 at 20:05

daniu

7,42521635

add a comment |

To use the short circuit but keep it more readable, you can do

boolean flag = method1();

flag = flag && method2();

flag = flag && method3();

And so on. Note you can't use flag &= methodX() because that doesn't short circuit.

edited Dec 21 '18 at 21:30

answered Dec 21 '18 at 20:05

daniu

7,42521635

To use the short circuit but keep it more readable, you can do

boolean flag = method1();

flag = flag && method2();

flag = flag && method3();

And so on. Note you can't use flag &= methodX() because that doesn't short circuit.

edited Dec 21 '18 at 21:30

answered Dec 21 '18 at 20:05

daniu

7,42521635

edited Dec 21 '18 at 21:30

answered Dec 21 '18 at 20:05

daniu

7,42521635

answered Dec 21 '18 at 20:05

daniu

7,42521635

answered Dec 21 '18 at 20:05

daniu

7,42521635

add a comment |

So first step is to create a method as such:

public static boolean apply(BooleanSupplier... funcs){

      return Arrays.stream(funcs).allMatch(BooleanSupplier::getAsBoolean);

}

This function consumes any number of functions that have the signature () -> boolean i.e. a supplier (a function taking no inputs and returns a boolean result).

Then you can call it as follows using a method reference:

if(apply(Main::method1, 

         Main::method2, 

         Main::method3, 

         Main::method4)){ ... };

Where Main is the class where the methods are defined.

or lambda:

if(apply(() -> method1(), 

         () -> method2(), 

         () -> method3(), 

         () -> method4())){ ... };

edited Dec 21 '18 at 22:44

answered Dec 21 '18 at 22:27

Aomine

41.9k74071

This one is likely more readable if the list is more than 3 or 4.

– jpmc26
Dec 22 '18 at 1:18

add a comment |

So first step is to create a method as such:

public static boolean apply(BooleanSupplier... funcs){

      return Arrays.stream(funcs).allMatch(BooleanSupplier::getAsBoolean);

}

This function consumes any number of functions that have the signature () -> boolean i.e. a supplier (a function taking no inputs and returns a boolean result).

Then you can call it as follows using a method reference:

if(apply(Main::method1, 

         Main::method2, 

         Main::method3, 

         Main::method4)){ ... };

Where Main is the class where the methods are defined.

or lambda:

if(apply(() -> method1(), 

         () -> method2(), 

         () -> method3(), 

         () -> method4())){ ... };

edited Dec 21 '18 at 22:44

answered Dec 21 '18 at 22:27

Aomine

41.9k74071

This one is likely more readable if the list is more than 3 or 4.

– jpmc26
Dec 22 '18 at 1:18

add a comment |

So first step is to create a method as such:

public static boolean apply(BooleanSupplier... funcs){

      return Arrays.stream(funcs).allMatch(BooleanSupplier::getAsBoolean);

}

This function consumes any number of functions that have the signature () -> boolean i.e. a supplier (a function taking no inputs and returns a boolean result).

Then you can call it as follows using a method reference:

if(apply(Main::method1, 

         Main::method2, 

         Main::method3, 

         Main::method4)){ ... };

Where Main is the class where the methods are defined.

or lambda:

if(apply(() -> method1(), 

         () -> method2(), 

         () -> method3(), 

         () -> method4())){ ... };

edited Dec 21 '18 at 22:44

answered Dec 21 '18 at 22:27

Aomine

41.9k74071

So first step is to create a method as such:

public static boolean apply(BooleanSupplier... funcs){

      return Arrays.stream(funcs).allMatch(BooleanSupplier::getAsBoolean);

}

This function consumes any number of functions that have the signature () -> boolean i.e. a supplier (a function taking no inputs and returns a boolean result).

Then you can call it as follows using a method reference:

if(apply(Main::method1, 

         Main::method2, 

         Main::method3, 

         Main::method4)){ ... };

Where Main is the class where the methods are defined.

or lambda:

if(apply(() -> method1(), 

         () -> method2(), 

         () -> method3(), 

         () -> method4())){ ... };

edited Dec 21 '18 at 22:44

answered Dec 21 '18 at 22:27

Aomine

41.9k74071

edited Dec 21 '18 at 22:44

answered Dec 21 '18 at 22:27

Aomine

41.9k74071

answered Dec 21 '18 at 22:27

Aomine

41.9k74071

answered Dec 21 '18 at 22:27

Aomine

41.9k74071

This one is likely more readable if the list is more than 3 or 4.

– jpmc26
Dec 22 '18 at 1:18

add a comment |

This one is likely more readable if the list is more than 3 or 4.

– jpmc26
Dec 22 '18 at 1:18

This one is likely more readable if the list is more than 3 or 4.

– jpmc26
Dec 22 '18 at 1:18

add a comment |

Nested conditional operators would solve this:

flag = method1() ? (method2() ? method3(): false): false;

answered Dec 21 '18 at 19:50

nicomp

2,60621231

nice +1, you can simplify it to flag = method1() && (method2() && method3());. assuming flag is needed after the condtions then this is a good approach.

– Aomine
Dec 21 '18 at 19:51

@Aomine I think that would be better answer than this

– eis
Dec 21 '18 at 19:52

@Aomine but the conditional operator get so little respect! :)

– nicomp
Dec 21 '18 at 19:53

1

@nicomp right, i'll post it as an answer then :). #eis sure thing.

– Aomine
Dec 21 '18 at 19:54

add a comment |

Nested conditional operators would solve this:

flag = method1() ? (method2() ? method3(): false): false;

answered Dec 21 '18 at 19:50

nicomp

2,60621231

nice +1, you can simplify it to flag = method1() && (method2() && method3());. assuming flag is needed after the condtions then this is a good approach.

– Aomine
Dec 21 '18 at 19:51

@Aomine I think that would be better answer than this

– eis
Dec 21 '18 at 19:52

@Aomine but the conditional operator get so little respect! :)

– nicomp
Dec 21 '18 at 19:53

1

@nicomp right, i'll post it as an answer then :). #eis sure thing.

– Aomine
Dec 21 '18 at 19:54

add a comment |

Nested conditional operators would solve this:

flag = method1() ? (method2() ? method3(): false): false;

answered Dec 21 '18 at 19:50

nicomp

2,60621231

Nested conditional operators would solve this:

flag = method1() ? (method2() ? method3(): false): false;

answered Dec 21 '18 at 19:50

nicomp

2,60621231

answered Dec 21 '18 at 19:50

nicomp

2,60621231

answered Dec 21 '18 at 19:50

nicomp

2,60621231

answered Dec 21 '18 at 19:50

nicomp

2,60621231

nice +1, you can simplify it to flag = method1() && (method2() && method3());. assuming flag is needed after the condtions then this is a good approach.

– Aomine
Dec 21 '18 at 19:51

@Aomine I think that would be better answer than this

– eis
Dec 21 '18 at 19:52

@Aomine but the conditional operator get so little respect! :)

– nicomp
Dec 21 '18 at 19:53

1

@nicomp right, i'll post it as an answer then :). #eis sure thing.

– Aomine
Dec 21 '18 at 19:54

add a comment |

nice +1, you can simplify it to flag = method1() && (method2() && method3());. assuming flag is needed after the condtions then this is a good approach.

– Aomine
Dec 21 '18 at 19:51

@Aomine I think that would be better answer than this

– eis
Dec 21 '18 at 19:52

@Aomine but the conditional operator get so little respect! :)

– nicomp
Dec 21 '18 at 19:53

1

@nicomp right, i'll post it as an answer then :). #eis sure thing.

– Aomine
Dec 21 '18 at 19:54

nice +1, you can simplify it to flag = method1() && (method2() && method3());. assuming flag is needed after the condtions then this is a good approach.

– Aomine
Dec 21 '18 at 19:51

@Aomine I think that would be better answer than this

– eis
Dec 21 '18 at 19:52

@Aomine but the conditional operator get so little respect! :)

– nicomp
Dec 21 '18 at 19:53

@nicomp right, i'll post it as an answer then :). #eis sure thing.

– Aomine
Dec 21 '18 at 19:54

add a comment |

You can use the && operator :

boolean flag = method1() && method2() && method3() && method4() && ...;

Do whatever with flag.

answered Dec 21 '18 at 20:03

fastcodejava

24.2k19109162

add a comment |

You can use the && operator :

boolean flag = method1() && method2() && method3() && method4() && ...;

Do whatever with flag.

answered Dec 21 '18 at 20:03

fastcodejava

24.2k19109162

add a comment |

You can use the && operator :

boolean flag = method1() && method2() && method3() && method4() && ...;

Do whatever with flag.

answered Dec 21 '18 at 20:03

fastcodejava

24.2k19109162

You can use the && operator :

boolean flag = method1() && method2() && method3() && method4() && ...;

Do whatever with flag.

answered Dec 21 '18 at 20:03

fastcodejava

24.2k19109162

answered Dec 21 '18 at 20:03

fastcodejava

24.2k19109162

answered Dec 21 '18 at 20:03

fastcodejava

24.2k19109162

answered Dec 21 '18 at 20:03

fastcodejava

24.2k19109162

add a comment |

incase the methods truth values have to be saved or other side-effects have to happen:

boolean flag;

switch(1){

  case 1: flag = method1();  

          //additional code goes here



          if(!flag) break; 

  case 2: flag = method2();

          //additional code goes here



          if(!flag) break; 

     ....

  default: flag = method10();

          if(!flag) break; 

}

you shouldn't want this and it's... unexpected.
but this is a readable way of adding additional statements to each code path.

otherwise: do it differently or fall back on ifs.

answered Dec 21 '18 at 21:13

RecurringError

623

add a comment |

incase the methods truth values have to be saved or other side-effects have to happen:

boolean flag;

switch(1){

  case 1: flag = method1();  

          //additional code goes here



          if(!flag) break; 

  case 2: flag = method2();

          //additional code goes here



          if(!flag) break; 

     ....

  default: flag = method10();

          if(!flag) break; 

}

you shouldn't want this and it's... unexpected.
but this is a readable way of adding additional statements to each code path.

otherwise: do it differently or fall back on ifs.

answered Dec 21 '18 at 21:13

RecurringError

623

add a comment |

incase the methods truth values have to be saved or other side-effects have to happen:

boolean flag;

switch(1){

  case 1: flag = method1();  

          //additional code goes here



          if(!flag) break; 

  case 2: flag = method2();

          //additional code goes here



          if(!flag) break; 

     ....

  default: flag = method10();

          if(!flag) break; 

}

you shouldn't want this and it's... unexpected.
but this is a readable way of adding additional statements to each code path.

otherwise: do it differently or fall back on ifs.

answered Dec 21 '18 at 21:13

RecurringError

623

incase the methods truth values have to be saved or other side-effects have to happen:

boolean flag;

switch(1){

  case 1: flag = method1();  

          //additional code goes here



          if(!flag) break; 

  case 2: flag = method2();

          //additional code goes here



          if(!flag) break; 

     ....

  default: flag = method10();

          if(!flag) break; 

}

you shouldn't want this and it's... unexpected.
but this is a readable way of adding additional statements to each code path.

otherwise: do it differently or fall back on ifs.

boolean flag;

switch(1){

  case 1: flag = method1();  

          //additional code goes here



          if(!flag) break; 

  case 2: flag = method2();

          //additional code goes here



          if(!flag) break; 

     ....

  default: flag = method10();

          if(!flag) break; 

}

boolean flag;

switch(1){

  case 1: flag = method1();  

          //additional code goes here



          if(!flag) break; 

  case 2: flag = method2();

          //additional code goes here



          if(!flag) break; 

     ....

  default: flag = method10();

          if(!flag) break; 

}

answered Dec 21 '18 at 21:13

RecurringError

623

answered Dec 21 '18 at 21:13

RecurringError

623

answered Dec 21 '18 at 21:13

RecurringError

623

answered Dec 21 '18 at 21:13

RecurringError

623

add a comment |

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