Why the second output is different??🤔

Question

https://code.sololearn.com/cEOUjwBEkfSA/?ref=app

So, first line we get "Hi 1" as output

Then, why in second line it shows "Hi 5"??

Shouldn't it show "Hi 5 1" or "Hi 1"??

Edit- I have edited my code

Answer 1

In your question values printing in the bases of priority
int x=5;
cout<<"Hi "<<(x && true)<<endl;

Here x is decleared as 5 which is positive integer so every positive number with condition treat as true either 1 so first here in first cout statement u y used brackets so brac will solve first and result will be Hi 1 .

But
cout<<"Hi "<<x && true;

Here u removed brac and << this extraction operator has highest priority so its printing 5 and 1
And work as multiply 5*1 and or as a plus so result is 5

Answer 2

A.S. I agree with your analysis up until this nonsense statement:
"... so its printing 5 and 1
And work as multiply 5*1 and or as a plus so result is 5"

Let's look again at the line
cout<<"Hi "<<x && true;

This is evaluated as a Boolean expression with two operands,
<A> && <B>;

First it evaluates <A>, which is
   cout<<"Hi "<<x.
The output is "Hi 5", and <A> evaluates to a non-null pointer. For Boolean purposes the pointer reduces to true.

Then it evaluates <B>, which is simply
   true.

Finally, <A>&&<B>, i.e., true&&true, reduces to true.

There is no further action to take with the expression so execution proceeds to the next statement without using the final true result.

Answer 3

A.S. 🤔 It's complicated but thanks 🙂

Answer 4

A.S. As you said, every positive number treated as true .. then why y && true is treated as true??

And false is equeals to 0..
Then x && false should be 0 or y && false should be 0 but why it shows 1??

Answer 5

Samael Morningstar read about logical operators you will understood better .

5&&true will give true here 5 is positive use bracket when u print values

5&&0 will give 0 or false

5||0 will give true ==1