Why the final value of b is 0?

Question

Can someone explain why this code outputs 0 (b equals zero)?
int main() {
int b = 0;
b = b++;
cout << b;
return 0;
}

Answer 1

what happens when you
x = x++;

is (assume u r passing x to a function)

function(x){
xOLD = x;
x = x+1;
return xOLD;
}

so,
x = x++;
becomes
x = (store old value of x --> do (x+1)--> return old value of x)
and hence,
x = x; (or x = xOLD)

For other variables it works perfectly,
y = x++;
becomes,
y = (store OLD value of x --> do (x+1) --> return OLD value of x);
and thus,
y=x;(or y = xOLD)

hope this clarifies your problem, thumbs up if it does👍

Answer 2

congrats wojciech on 500 challenge victories...amazing

Answer 3

@gowtham
The single line
b = b++;

becomes

temp = b;
b = b+1;
b = temp;

for others, eg
y = b++;

expands to;

temp = b;
b = b+1;
y = temp;

Answer 4

It should be

case 1{
int b = 0;
b = b++; //b=0
}

case 2{
int b = 0;
b = ++b; //b=1
}

case 3{
int a = 0;
int b = 0;
a = b++; //a=0 and b=1
}

case 4{
int a = 0;
int b = 0;
a = ++b; //a=1 and b=1
}

Answer 5

ok, but isn't this single line equal to two lines:
b = b;
b++
?
It's a bit confusing for me. That is why I try to avoid such constructions in my code.

Answer 6

@paul kabira
then in your function how would you access the new "x" as u said it returns xOLD variable value.

Answer 7

b++ return b so 0 so b = 0

Answer 8

when u write b = ++b;
the result be : 1

Answer 9

Read the course.
b++ returns the currwnt value of b, THEN adds 1

Answer 10

it is post increment operation so the value of variable is updated afterwards

Answer 11

prefix and postfix operator
++b is prefix...it will increment the value then use it.
but in case of b++.
first it will use old value then increment it.
so b= b++ . cout<< b gives 0

Answer 12

since it is a post increment operator ,the previous value of b that is '0' assigned to b again then it will get incremented...
so the value of b remains same...
therefore output is 0