anyone who can tell me?

[Jin chengong]

here is the code:
public class Inc {
public static void main(String[] args) {
Inc inc = new Inc();
int i = 0;
inc.fermin(i);
i = i++;
System.out.println(i);
}
void fermin(int i){
i++;
}
}

the answer is 0 not 1.why ?

[Remko Strating]

i++ is a postfix operator so first a value is assigned and then the value will be increased by one.
If you use a prefix operator the result will be as you will expect.

public class Test {
public static void main(String[] args) {
Test test = new Test();
int i = 0;
test.fermin(i);
i = ++i;
System.out.println(i);
}
void fermin(int i){
i++;
}
}

For more information over postfix operators see the following link


Operators and Assignments - Prefix and Postfix Operators

[Rob Spoor]

inc.fermin(i);

This call works on a local variable that is initialized with the value of i, 0. When in that method i++ is called, it increments that local variable, not the variable passed to the method.

i = i++;

Here's the real "magic". What happens here is something like the following:

int temp = i; // first i is evaluated, temp == 0
i = i + 1; // i is incremented, i == 1
i = temp; // the assignment is made, i == 0 again

Were you to write i++ it would have been 1. Had you written "i = ++i" it would have worked as well:

i = i + 1;
int temp = i;
i = temp;

For more information, search for the difference between prefix and postfix increment.

[Jin chengong]

think you very much!

[Jin chengong]

Thanks for the reply.
Remko Strating
Rob Prime