final instance variable can be reassigned to some other value using const in dart
Does anyone know why the below code works in dart. final keyword is used to define constant variables. But the below code works little different. If we are using const with a different value it's working fine without giving error.
void main() {
ExampleFinal exampleFinal = new ExampleFinal();
}
class ExampleFinal() {
final a = 5;
ExampleFinal() {
// The below statement will not create any error.
// But if you are remove const in below line it'll show a compile time error.
const a = 6;
print(a); // Prints 6
}
}
Is it a bug or a feature in dart? There is nothing like mentioned in document also.
dart
asked Jan 1 at 17:45
Mahendran SakkaraiMahendran Sakkarai
5,54232956
add a comment |
Does anyone know why the below code works in dart. final keyword is used to define constant variables. But the below code works little different. If we are using const with a different value it's working fine without giving error.
void main() {
ExampleFinal exampleFinal = new ExampleFinal();
}
class ExampleFinal() {
final a = 5;
ExampleFinal() {
// The below statement will not create any error.
// But if you are remove const in below line it'll show a compile time error.
const a = 6;
print(a); // Prints 6
}
}
Is it a bug or a feature in dart? There is nothing like mentioned in document also.
dart
asked Jan 1 at 17:45
Mahendran SakkaraiMahendran Sakkarai
5,54232956
add a comment |
Does anyone know why the below code works in dart. final keyword is used to define constant variables. But the below code works little different. If we are using const with a different value it's working fine without giving error.
void main() {
ExampleFinal exampleFinal = new ExampleFinal();
}
class ExampleFinal() {
final a = 5;
ExampleFinal() {
// The below statement will not create any error.
// But if you are remove const in below line it'll show a compile time error.
const a = 6;
print(a); // Prints 6
}
}
Is it a bug or a feature in dart? There is nothing like mentioned in document also.
dart
asked Jan 1 at 17:45
Mahendran SakkaraiMahendran Sakkarai
5,54232956
Does anyone know why the below code works in dart. final keyword is used to define constant variables. But the below code works little different. If we are using const with a different value it's working fine without giving error.
void main() {
ExampleFinal exampleFinal = new ExampleFinal();
}
class ExampleFinal() {
final a = 5;
ExampleFinal() {
// The below statement will not create any error.
// But if you are remove const in below line it'll show a compile time error.
const a = 6;
print(a); // Prints 6
}
}
Is it a bug or a feature in dart? There is nothing like mentioned in document also.
dart
dart
asked Jan 1 at 17:45
Mahendran SakkaraiMahendran Sakkarai
5,54232956
asked Jan 1 at 17:45
Mahendran SakkaraiMahendran Sakkarai
5,54232956
edited Jan 1 at 17:47
Mahendran Sakkarai
asked Jan 1 at 17:45
Mahendran SakkaraiMahendran Sakkarai
5,54232956
asked Jan 1 at 17:45
Mahendran SakkaraiMahendran Sakkarai
5,54232956
asked Jan 1 at 17:45
Mahendran SakkaraiMahendran Sakkarai
5,54232956
5,54232956
add a comment |
add a comment |
1 Answer
1
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votes
const a = 6;
creates a new variable that shadows the final a = 5;
This is possible because {...} creates a new scope in the constructor body.
If you add at the end of the constructor
print(this.a);
it will print 5
answered Jan 1 at 17:47
Günter ZöchbauerGünter Zöchbauer
327k69980921
Looks interesting. But it may confuse also. And there is no such thing is mentioned in the document also!
– Mahendran Sakkarai
Jan 1 at 18:02
1
Haven't checked the docs for a while. A type and/orvar,const,finalbefore a variable always create a new one. In the same scope you get an error if there are name conflicts. Most languages I worked with behave very similar.
– Günter Zöchbauer
Jan 1 at 18:08
add a comment |
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1 Answer
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1 Answer
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active
oldest
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const a = 6;
creates a new variable that shadows the final a = 5;
This is possible because {...} creates a new scope in the constructor body.
If you add at the end of the constructor
print(this.a);
it will print 5
answered Jan 1 at 17:47
Günter ZöchbauerGünter Zöchbauer
327k69980921
Looks interesting. But it may confuse also. And there is no such thing is mentioned in the document also!
– Mahendran Sakkarai
Jan 1 at 18:02
1
Haven't checked the docs for a while. A type and/orvar,const,finalbefore a variable always create a new one. In the same scope you get an error if there are name conflicts. Most languages I worked with behave very similar.
– Günter Zöchbauer
Jan 1 at 18:08
add a comment |
const a = 6;
creates a new variable that shadows the final a = 5;
This is possible because {...} creates a new scope in the constructor body.
If you add at the end of the constructor
print(this.a);
it will print 5
answered Jan 1 at 17:47
Günter ZöchbauerGünter Zöchbauer
327k69980921
Looks interesting. But it may confuse also. And there is no such thing is mentioned in the document also!
– Mahendran Sakkarai
Jan 1 at 18:02
1
Haven't checked the docs for a while. A type and/orvar,const,finalbefore a variable always create a new one. In the same scope you get an error if there are name conflicts. Most languages I worked with behave very similar.
– Günter Zöchbauer
Jan 1 at 18:08
add a comment |
const a = 6;
creates a new variable that shadows the final a = 5;
This is possible because {...} creates a new scope in the constructor body.
If you add at the end of the constructor
print(this.a);
it will print 5
answered Jan 1 at 17:47
Günter ZöchbauerGünter Zöchbauer
327k69980921
const a = 6;
creates a new variable that shadows the final a = 5;
This is possible because {...} creates a new scope in the constructor body.
If you add at the end of the constructor
print(this.a);
it will print 5
answered Jan 1 at 17:47
Günter ZöchbauerGünter Zöchbauer
327k69980921
answered Jan 1 at 17:47
Günter ZöchbauerGünter Zöchbauer
327k69980921
answered Jan 1 at 17:47
Günter ZöchbauerGünter Zöchbauer
327k69980921
answered Jan 1 at 17:47
Günter ZöchbauerGünter Zöchbauer
327k69980921
327k69980921
Looks interesting. But it may confuse also. And there is no such thing is mentioned in the document also!
– Mahendran Sakkarai
Jan 1 at 18:02
1
Haven't checked the docs for a while. A type and/orvar,const,finalbefore a variable always create a new one. In the same scope you get an error if there are name conflicts. Most languages I worked with behave very similar.
– Günter Zöchbauer
Jan 1 at 18:08
add a comment |
Looks interesting. But it may confuse also. And there is no such thing is mentioned in the document also!
– Mahendran Sakkarai
Jan 1 at 18:02
1
Haven't checked the docs for a while. A type and/orvar,const,finalbefore a variable always create a new one. In the same scope you get an error if there are name conflicts. Most languages I worked with behave very similar.
– Günter Zöchbauer
Jan 1 at 18:08
Looks interesting. But it may confuse also. And there is no such thing is mentioned in the document also!
– Mahendran Sakkarai
Jan 1 at 18:02
Looks interesting. But it may confuse also. And there is no such thing is mentioned in the document also!
– Mahendran Sakkarai
Jan 1 at 18:02
1
1
Haven't checked the docs for a while. A type and/or
var, const, final before a variable always create a new one. In the same scope you get an error if there are name conflicts. Most languages I worked with behave very similar.– Günter Zöchbauer
Jan 1 at 18:08
Haven't checked the docs for a while. A type and/or
var, const, final before a variable always create a new one. In the same scope you get an error if there are name conflicts. Most languages I worked with behave very similar.– Günter Zöchbauer
Jan 1 at 18:08
add a comment |
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