about cout cin 2 var

Question

if i type

int a,b;
cin >> a,b;
cout << “a and b = ”<<a <<“and”<<b;

if i input a=1 b=2
why output are 1 and 0?

Answer 1

To get more than one variable using cin, it should look like
cin >> a >> b;
In your case, b is not being initialized, hence it still has some garbage value, in this case 0.

Answer 2

Ipang, do not take me for granted here, but I think that since the comma operator always has the lowest precedence, first
cin >> a
is called, afterwards the result is evaluated and discarded by the comma operator, and then b is evaluated and returned. Try running the following code snippet:

int a { 1 }, b { 2 };

auto c = ( cin >> a, b );

cout << a << endl
<< b << endl
<< c << endl;

Here c will equal b, because ultimately, b will be returned from the expression. Note the neccessary braces, without them the assignment is done before the comma operator is evaluated, resulting in c being assigned the result from
cin >> a
and therefore resulting in a compilation error.

Answer 3

Ipang, sorry for the misunderstanding, "do not take me for granted" <=> "do not assume me to be 100% correct on this, better do own research additionally". The correct term should have been "do not take this at face value". I wouldn't ever be annoyed by a ping of yours. :)
Anyway, a compilation of pages that might be helpful, although you probably know most of them already:

https://en.cppreference.com/w/cpp/language/operator_other

https://www.fluentcpp.com/2018/07/31/how-to-get-along-with-the-comma-operator/

https://stackoverflow.com/questions/54142/how-does-the-comma-operator-work

https://www.geeksforgeeks.org/comna-in-c-and-c/

https://codassium.com/blog/the-forgotten-comma-operator/

Answer 4

Shadow
I was wondering how comma operator behaviour in that case, still not so clear about comma operator behaviour in different occasions. Need a little help here 😁

Answer 5

Shadow
First, I do not take you for granted - I hope my ping didn't bother you in any way.

Thanks for replying, I asked because I found a lot of codes in Playground written to cover the wonder of this operator, and to see those codes kinda made me feel unsure about how the operator behaves, because what I predict doesn't match the code output sometimes (read: my prediction of code output be wrong). I'll see to it that I can solve this soon ✌

Answer 6

No problem at all 😁
And Big Thanks for the references Shadow, I'll get to them soon. This really helps Thanks again 👍

Answer 7

oh i c thx~