Why b=b++ does not change b value?

Question

int b=1;
b=b++;
cout<<b;
Result will be 1. Why b !=2 ?
int x=b++;
cout<<endl<<b;
Result will be 2

Answer 1

because it first store b in b variable then add 1. You should do b = ++b; in order to add 1 to b then store it on b.

Answer 2

Thanks for the reply, but I know how ++b works.
I'm interested in b=b++
I think it works like that
1) b value pushed to stack
2) b value incremented by 1
3) old b value poped from stack into b.
That is why b is still the same.

Other words, It is similar to
int temp = b;
b++;
b=temp;

Answer 3

Typically post increment is implemented as :
int post++(int &i) {
int temp = *i;
*i = *i + 1;
return temp;
}
We can observe here that when we use
b= b++ then first the value of b is incremented at the original location and returns the previous value of b, so it becomes
b= 1, as the incremented value is overridden.

Answer 4

if you use b++, you need to recall b++ in a loop, since the value changes after it has been set. in your case, b is declared but it never gets triggered, therefore it stays 1.
in the second example you call on b++, therefore it triggers the increment of b