Java Personal learning Note self-increment (pre-and post) assignment operation (extended assignment operation) logical operator

Source: Internet
Author: User

+ + self-increment front: First operation after the value a=3; B=++a; a=4; b=4;

+ + + self-increment: First take value after Operation A=3; b=a++; a=4; b=3;

--self-reducing pre-set: First operation after the value a=3; b=--a;a=2;b=2;

--Self-reduction after: the first value after the Operation A=3;b=a--;a=2;b=3;

The preceding is the first operation after the value is set after the value of the operation.

Assignment Operation +

Extended Assignment Operation + =,-+, *=,/=,%=

Short S = 3;

s=s+2; ①

s+=2; Ii

What's the difference between ① and ②?

1 is the compilation does not pass, because s = S +2, the integer 2 in Java by default is the type int, S + 2 will first be converted to the int type to calculate, obtained an int type result. Then the int type cannot be converted directly to short, so there is a compilation error.

2 is correct, + = is an extended assignment operation, so the essence is assignment. The assignment operation does not change the data type. In fact, what s+= do is to get the result of S + 2 first 5, and then assign the value 5 to S.

logical operators

& Logic and | Logical OR! Logical Non-

&& Short circuit with | | Short circuit or ^ logic XOR

"&" regardless of whether the left is true or false, the right side will be calculated. "&&" When the left is false, the right is no longer counted.

"|" Whether or not the left is true or false, the right side will be calculated. "| |" When the left is true, the right side is no longer counted.

^ xor or when both conditions are true or both are false, the result is false. The pursuit is "different".

int x = 1;

int Y=1;

if (x++==2 & ++y==2) {

x = 7;

}

System.out.println ("x=" +x+ ", y=" +y);

The result above is x=2,y=2 interpretation & operation, regardless of whether the left is true or false, the right side will participate in the operation. However, the rear-facing + + is the first value after the operation. That is, x + = = 2 is not true, at this time the X. 1

int x = 1,y = 1;

if (x++==2 && ++y==2) {

x = 7;

}

System.out.println ("x=" +x+ ", y=" +y);

The result above is x=2,y=1 because the && operation on the left is false and the right side is no longer involved in the operation. So the result is x=2 y=1 (++y = = 2 Not participating in the operation)

int x = 1,y = 1;

if (x++==1 | ++y==1) {

x = 7;

}

System.out.println ("x=" +x+ ", y=" +y);

The result above is x=7,y=2 because | operations, regardless of the left is true false right are involved in the operation, so ++y = = 1 to participate in the operation. At this point the if condition is set to execute x = 7.

int x = 1,y = 1;

if (x++==1 | | ++y==1) {

x = 7;

}

System.out.println ("x=" +x+ ", y=" +y);

The result above is X=7,y=1, and X + + is back-mounted. The value is evaluated first. So x + + = 1 was established. Because | | Operation, when the left is true, the right side is no longer involved in the operation. So ++y==1 does not execute. However, the operation has already been established. So x=7, Y=1.

Java Personal learning Note self-increment (pre-and post) assignment operation (extended assignment operation) logical operator

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