please explain the output?

Question

int b=3;
b=b++;
cout<<b;
the output is still 3!!

Answer 1

Well after a long argument I finally got the answer that this (multiple increment /decrement operator) is compiler dependent.

Answer 2

thanks for your answer @ali kamyanfar

Answer 3

++b icreases the value in first iteration. whereas b++ increase value by 1 but can be used in next time.or next step only.

Answer 4

sir...maybe...first it is incremented to 4 then add to 4...total 8...but there is post increment..that is why it get incremented to 9...++I(4 pre)+I++(still 4 )=8...but at the last there was post increment which still didn't get incremented..so after incrementation 9

Answer 5

thanks for your answer Aditya kumar pandey

Answer 6

for example
int x=3;
int x=x++;/* before 3 change to 4 and 1 add to x in x++ 3 move to x so x still is 3 */
x=++x;/* first 1 add to x in ++x then move to x so x=4 */
cout<<x;
//out put is : 4

@Rishabh Agrawal
@sagar

Answer 7

At first line value of b is assign to 3.
On second line it is b++. What does it mean. Increase value of a variable by 1 but if it used it again. Means if It use again its value increase. Now you assign b++ value to b. So, what is value of b now its 3.
On 4th line its value is output. What is value of b. Is it 4, no. It is 3. because you assign value of b++ to b and here b is used again but it already assign value to 3. Now after that there is nothing like increment. If b++; was written instead of b=b++. Than its output will 4.

Answer 8

b=b++ does three things. Expression evaluation, increment, and assignment. In which order?

Since it is a post-increment, the increment is executed after the expression evaluation (thus b is assigned the original value of b, that is 3). In principle the increment could be performed also after the assignment, but as this contrived example shows that is not the case (otherwise b would become 4).

Hence the order of operations is expression evaluation, increment, and assignment.

EDIT
I tested the code in the code playground section of the C++ course and the output is actually 3, so unless it is not working properly the OP appears to be mistaken.
Revising my reasoning, the order of operations is evaluation, assignment, and increment.

Answer 9

OutPut in your questions is wrong..that should be 4...

Answer 10

its value is 4 but print 3 then add 1 to b .
but in ++b its value is 4 and print 4.

Answer 11

that means if i display b again it will show 4.
@Ali kamyanfar

Answer 12

@Ali kamyanfar
can u explain me what is exactly happening at 2 nd like of code as by setting the code, the variable is being incremented twice but output is only after 1 increment.
It will be good if u could explain it to me in detail.

Answer 13

int main(){
int b=3;
b=b++;/*before 3 change to 4 move to b so b still is 3 */
b=b++; //b still is 3
b=b++; //b still is 3
b =++b; /* b is 4 now because 3 change to 4 then move to b */
cout <<b;
//out put is 4
}

Answer 14

in the program the value assigned to 'b' is 3 and then the 3 is incremented to 4, but the value now stored in b remains 3

to change the value of 'b' from 3 to 4, two things can be done

1. b=++b:

2. b++=b: