Why the first swap attempt works, but the second doesn't work? [duplicate]
This question already has an answer here:
Is Java “pass-by-reference” or “pass-by-value”?
76 answers
Java: Why does this swap method not work? [duplicate]
10 answers
package main;
public class Main {
double radius;
public Main(double newRadius) {
radius = newRadius;
}
public static void main (String args) {
Main x = new Main(1);
Main y = new Main(2);
Main temp;
// try to swap first time
temp = x;
x = y;
y = temp;
System.out.println(x.radius + " " + y.radius);
x = new Main(1);
y = new Main(2);
// try to swap second time
swap(x, y);
System.out.println(x.radius + " " + y.radius);
}
public static void swap(Main x, Main y) {
Main temp = x;
x = y;
y = temp;
}
}
Why the first time worked, but the second time didn't? The first one did the swap, but the second one didn't. I'm passing the reference to the function. Why this doesn't work?
java oop
asked Nov 10 at 13:09
Roy Derek
6
marked as duplicate by Mureinik java
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Nov 10 at 13:12
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.
add a comment |
This question already has an answer here:
Is Java “pass-by-reference” or “pass-by-value”?
76 answers
Java: Why does this swap method not work? [duplicate]
10 answers
package main;
public class Main {
double radius;
public Main(double newRadius) {
radius = newRadius;
}
public static void main (String args) {
Main x = new Main(1);
Main y = new Main(2);
Main temp;
// try to swap first time
temp = x;
x = y;
y = temp;
System.out.println(x.radius + " " + y.radius);
x = new Main(1);
y = new Main(2);
// try to swap second time
swap(x, y);
System.out.println(x.radius + " " + y.radius);
}
public static void swap(Main x, Main y) {
Main temp = x;
x = y;
y = temp;
}
}
Why the first time worked, but the second time didn't? The first one did the swap, but the second one didn't. I'm passing the reference to the function. Why this doesn't work?
java oop
asked Nov 10 at 13:09
Roy Derek
6
marked as duplicate by Mureinik java
Users with the java badge can single-handedly close java questions as duplicates and reopen them as needed.
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Nov 10 at 13:12
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.
5
References are passed by value. The variables inside the method are not the variables outside the method, so reassigning them doesn't alter external variables. See Is Java pass-by-reference or pass-by-value?
– khelwood
Nov 10 at 13:11
add a comment |
This question already has an answer here:
Is Java “pass-by-reference” or “pass-by-value”?
76 answers
Java: Why does this swap method not work? [duplicate]
10 answers
package main;
public class Main {
double radius;
public Main(double newRadius) {
radius = newRadius;
}
public static void main (String args) {
Main x = new Main(1);
Main y = new Main(2);
Main temp;
// try to swap first time
temp = x;
x = y;
y = temp;
System.out.println(x.radius + " " + y.radius);
x = new Main(1);
y = new Main(2);
// try to swap second time
swap(x, y);
System.out.println(x.radius + " " + y.radius);
}
public static void swap(Main x, Main y) {
Main temp = x;
x = y;
y = temp;
}
}
Why the first time worked, but the second time didn't? The first one did the swap, but the second one didn't. I'm passing the reference to the function. Why this doesn't work?
java oop
asked Nov 10 at 13:09
Roy Derek
6
This question already has an answer here:
Is Java “pass-by-reference” or “pass-by-value”?
76 answers
Java: Why does this swap method not work? [duplicate]
10 answers
package main;
public class Main {
double radius;
public Main(double newRadius) {
radius = newRadius;
}
public static void main (String args) {
Main x = new Main(1);
Main y = new Main(2);
Main temp;
// try to swap first time
temp = x;
x = y;
y = temp;
System.out.println(x.radius + " " + y.radius);
x = new Main(1);
y = new Main(2);
// try to swap second time
swap(x, y);
System.out.println(x.radius + " " + y.radius);
}
public static void swap(Main x, Main y) {
Main temp = x;
x = y;
y = temp;
}
}
Why the first time worked, but the second time didn't? The first one did the swap, but the second one didn't. I'm passing the reference to the function. Why this doesn't work?
This question already has an answer here:
Is Java “pass-by-reference” or “pass-by-value”?
76 answers
Java: Why does this swap method not work? [duplicate]
10 answers
java oop
java oop
asked Nov 10 at 13:09
Roy Derek
6
asked Nov 10 at 13:09
Roy Derek
6
asked Nov 10 at 13:09
Roy Derek
6
asked Nov 10 at 13:09
Roy Derek
6
asked Nov 10 at 13:09
Roy Derek
6
6
marked as duplicate by Mureinik java
Users with the java badge can single-handedly close java questions as duplicates and reopen them as needed.
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Nov 10 at 13:12
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.
marked as duplicate by Mureinik java
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Nov 10 at 13:12
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.
5
References are passed by value. The variables inside the method are not the variables outside the method, so reassigning them doesn't alter external variables. See Is Java pass-by-reference or pass-by-value?
– khelwood
Nov 10 at 13:11
add a comment |
5
References are passed by value. The variables inside the method are not the variables outside the method, so reassigning them doesn't alter external variables. See Is Java pass-by-reference or pass-by-value?
– khelwood
Nov 10 at 13:11
5
5
References are passed by value. The variables inside the method are not the variables outside the method, so reassigning them doesn't alter external variables. See Is Java pass-by-reference or pass-by-value?
– khelwood
Nov 10 at 13:11
References are passed by value. The variables inside the method are not the variables outside the method, so reassigning them doesn't alter external variables. See Is Java pass-by-reference or pass-by-value?
– khelwood
Nov 10 at 13:11
add a comment |
1 Answer
1
active
oldest
votes
Your mistaking how references are passed, your making a scope where references are swapped and then that scope ends.
Try storing the value of the field in the variable e.g. temp = x.radius and then assign to y.radius.
The reason why it works the first time is that the scope is the same.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your mistaking how references are passed, your making a scope where references are swapped and then that scope ends.
Try storing the value of the field in the variable e.g. temp = x.radius and then assign to y.radius.
The reason why it works the first time is that the scope is the same.
add a comment |
Your mistaking how references are passed, your making a scope where references are swapped and then that scope ends.
Try storing the value of the field in the variable e.g. temp = x.radius and then assign to y.radius.
The reason why it works the first time is that the scope is the same.
add a comment |
Your mistaking how references are passed, your making a scope where references are swapped and then that scope ends.
Try storing the value of the field in the variable e.g. temp = x.radius and then assign to y.radius.
The reason why it works the first time is that the scope is the same.
Your mistaking how references are passed, your making a scope where references are swapped and then that scope ends.
Try storing the value of the field in the variable e.g. temp = x.radius and then assign to y.radius.
The reason why it works the first time is that the scope is the same.
edited Nov 10 at 14:33
community wiki
2 revs
Jay
add a comment |
add a comment |
5
References are passed by value. The variables inside the method are not the variables outside the method, so reassigning them doesn't alter external variables. See Is Java pass-by-reference or pass-by-value?
– khelwood
Nov 10 at 13:11