Explain this code ......

Question

int main() {
int x = 0
cout << (--x) + (--x) ;
}

// output -4

how ??

Answer 1

Well... no you're wrong 😂
But you have understood.
However, when you write
y=(--x)+(--x)+(--x);

C++ thinks this :
y=(--x)+(--x);
y=y+(--x);

Soooo :
y=(--x)+(--x) ;// -2+-2=-4
y=y+(--x);// -4+-3=-7

--x //-1
--x //-2
x+x // x=-2
--x //-3
-2+x // x=-3

inctementation is fantastic but it becomes quick very very complicated as you can see.

If you don't understand ask me, don't hesitate!

Answer 2

that's a good question
--x changes x value before execute the statement.
So in this code we have :
(--x) // x= -1
(--x) // x= -2
AND HERE IS THE THING :
we don't have -1 + -2
we still have x + x // and x= -2
so -2 + -2 = -4

Ask if you don't understand

Answer 3

ya.. i was wrong :) but now
i have understood.

if we have four (--x) .
// cout << (--x) + (--x) + (--x) + (--x) ;

then

a = (--x) + (--x) //1st and 2nd (--x)
a = a + (--x) //3rd (--x)
a = a + (--x) //4th (--x)

so answer is

-11 :)

Answer 4

Exactly !
Now you can enjoy incrementation in C++ !

Answer 5

Thank you :)
for explanation

Answer 6

@jojo

if we have one more (--x) in COUT
// cout << (--x) + (--x) + (--x) ;
then x value is
(--x) // x = -1
(--x) // x = -2
(--x) // x = -3

and x+x+x // x = -3
so answer is -9

am i right ??