CODEWITHSUNDEEP
javasyntaxincrement
Will Sherwood
Will Sherwood

Reputation: 1484

java weird increment syntax

So I have always been taught that in Java, using the increment operator after a variable name in an expression will do the expression, and then increment the value and using the operator before a variable name in an expression will do the increment before the evaluation. Like this:

int x = 0;
int y = x++;

after this executes y should be 0 and x should be 1. and in this example

int x = 0;
int y = ++x;

should be x = 1 and y = 1.

Following that same logic, the following...

int x = 0;
int y = 0;
x = y++ - y++;

should output 0 as x and 2 as y because 0 - 0 = 0. However the output is

x = -1
y = 2

Why is this?

Edit: the value of y does not matter. x will always equal -1 and y will (in the end) equal y + 2.

Upvotes: 0

Views: 260

Answers (4)

Peshal
Peshal

Reputation: 1518

Looks like all you assumptions are incorrect. In your first case

int x = 0;
int y = x++;

x is 0 and y is 1 because y takes the incremented value of x. Similarly for your second case x is 0 and y is 1. For your final case:

int x = 0;
int y = 0;
x = y++(y is 1 here) - y++(y is 2 now);

so here y++, makes y 1 then you subtract this 1 with another increment of y than makes y 2 at that point so x = 1-2= -1

Upvotes: 0

WeaponsGrade
WeaponsGrade

Reputation: 878

I was taught that when there was a variable++ it always happened after the statement was done.

Not after the statement, but rather after the expression.

Any time you have n - (n + 1) you get the result -1. The right-hand side of an assignment must be fully evaluated before the assignment can take place.

Upvotes: 0

Scary Wombat
Scary Wombat

Reputation: 44813

x = y++ - y++;

equals 

int a = 0;
int b = 0;
int c = 0;
a = y++; // 0
b = y++; // 1
c = y;
-1 = a - b;
2 = y;

Upvotes: 0

Sotirios Delimanolis
Sotirios Delimanolis

Reputation: 279870

int x = 0;
int y = 0;
x = y++ - y++;
x = (0) - (1)
y = 1 ---> 2 // after ++

So x = -1 and y = 2

Upvotes: 2

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