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 ..
java
edited Dec 21 '18 at 20:00
Federico Peralta Schaffner
22k43370
asked Dec 21 '18 at 19:42
user10821509
624
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 ..
java
edited Dec 21 '18 at 20:00
Federico Peralta Schaffner
22k43370
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 returnflagafter 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 ..
java
edited Dec 21 '18 at 20:00
Federico Peralta Schaffner
22k43370
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
java
edited Dec 21 '18 at 20:00
Federico Peralta Schaffner
22k43370
asked Dec 21 '18 at 19:42
user10821509
624
edited Dec 21 '18 at 20:00
Federico Peralta Schaffner
22k43370
asked Dec 21 '18 at 19:42
user10821509
624
edited Dec 21 '18 at 20:00
Federico Peralta Schaffner
22k43370
edited Dec 21 '18 at 20:00
Federico Peralta Schaffner
22k43370
edited Dec 21 '18 at 20:00
Federico Peralta Schaffner
22k43370
22k43370
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
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 returnflagafter 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 returnflagafter 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
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
Do you need to return
flag after the last method has been executed?– Federico Peralta Schaffner
Dec 21 '18 at 19:50
1
1
"best" by what metric?
– TylerH
Dec 21 '18 at 22:52
"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).
answered Dec 21 '18 at 19:46
forpas
8,8441421
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()
answered Dec 21 '18 at 19:54
Aomine
40.3k73870
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.
answered Dec 21 '18 at 20:05
daniu
7,28021635
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())){ ... };
answered Dec 21 '18 at 22:27
Aomine
40.3k73870
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
2,51921230
nice +1, you can simplify it toflag = method1() && (method2() && method3());. assumingflagis 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
24k19109162
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
active
oldest
votes
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).
answered Dec 21 '18 at 19:46
forpas
8,8441421
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).
answered Dec 21 '18 at 19:46
forpas
8,8441421
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).
answered Dec 21 '18 at 19:46
forpas
8,8441421
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).
answered Dec 21 '18 at 19:46
forpas
8,8441421
edited Dec 21 '18 at 20:02
answered Dec 21 '18 at 19:46
forpas
8,8441421
answered Dec 21 '18 at 19:46
forpas
8,8441421
answered Dec 21 '18 at 19:46
forpas
8,8441421
8,8441421
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
That doesn't put anything in flag.
– nicomp
Dec 21 '18 at 19:47
8
8
The flag is not needed this way.
– forpas
Dec 21 '18 at 19:48
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
This is hacker way :) But makes code, probably, hard to read.
– Koziołek
Dec 21 '18 at 19:54
6
6
@Koziołek If you know what you're doing, it is not hard to read.
– forpas
Dec 21 '18 at 19:55
@Koziołek If you know what you're doing, it is not hard to read.
– forpas
Dec 21 '18 at 19:55
2
2
@Koziołek as I said If you know what you're doing....
– forpas
Dec 21 '18 at 20:01
@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()
answered Dec 21 '18 at 19:54
Aomine
40.3k73870
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()
answered Dec 21 '18 at 19:54
Aomine
40.3k73870
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()
answered Dec 21 '18 at 19:54
Aomine
40.3k73870
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()
answered Dec 21 '18 at 19:54
Aomine
40.3k73870
edited Dec 21 '18 at 20:00
answered Dec 21 '18 at 19:54
Aomine
40.3k73870
answered Dec 21 '18 at 19:54
Aomine
40.3k73870
answered Dec 21 '18 at 19:54
Aomine
40.3k73870
40.3k73870
add a comment |
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.
answered Dec 21 '18 at 20:05
daniu
7,28021635
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.
answered Dec 21 '18 at 20:05
daniu
7,28021635
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.
answered Dec 21 '18 at 20:05
daniu
7,28021635
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.
answered Dec 21 '18 at 20:05
daniu
7,28021635
edited Dec 21 '18 at 21:30
answered Dec 21 '18 at 20:05
daniu
7,28021635
answered Dec 21 '18 at 20:05
daniu
7,28021635
answered Dec 21 '18 at 20:05
daniu
7,28021635
7,28021635
add a comment |
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())){ ... };
answered Dec 21 '18 at 22:27
Aomine
40.3k73870
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 |
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())){ ... };
answered Dec 21 '18 at 22:27
Aomine
40.3k73870
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 |
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())){ ... };
answered Dec 21 '18 at 22:27
Aomine
40.3k73870
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())){ ... };
answered Dec 21 '18 at 22:27
Aomine
40.3k73870
edited Dec 21 '18 at 22:44
answered Dec 21 '18 at 22:27
Aomine
40.3k73870
answered Dec 21 '18 at 22:27
Aomine
40.3k73870
answered Dec 21 '18 at 22:27
Aomine
40.3k73870
40.3k73870
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
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
2,51921230
nice +1, you can simplify it toflag = method1() && (method2() && method3());. assumingflagis 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
2,51921230
nice +1, you can simplify it toflag = method1() && (method2() && method3());. assumingflagis 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
2,51921230
Nested conditional operators would solve this:
flag = method1() ? (method2() ? method3(): false): false;
answered Dec 21 '18 at 19:50
nicomp
2,51921230
answered Dec 21 '18 at 19:50
nicomp
2,51921230
answered Dec 21 '18 at 19:50
nicomp
2,51921230
answered Dec 21 '18 at 19:50
nicomp
2,51921230
2,51921230
nice +1, you can simplify it toflag = method1() && (method2() && method3());. assumingflagis 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 toflag = method1() && (method2() && method3());. assumingflagis 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
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 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
@Aomine but the conditional operator get so little respect! :)
– nicomp
Dec 21 '18 at 19:53
1
1
@nicomp right, i'll post it as an answer then :). #eis sure thing.
– Aomine
Dec 21 '18 at 19:54
@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
24k19109162
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
24k19109162
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
24k19109162
You can use the && operator :
boolean flag = method1() && method2() && method3() && method4() && ...;
Do whatever with flag.
answered Dec 21 '18 at 20:03
fastcodejava
24k19109162
answered Dec 21 '18 at 20:03
fastcodejava
24k19109162
answered Dec 21 '18 at 20:03
fastcodejava
24k19109162
answered Dec 21 '18 at 20:03
fastcodejava
24k19109162
24k19109162
add a comment |
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
623
add a comment |
add a comment |
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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
flagafter 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