why the output of this code equal 1 ?

Question

int x=10,y=20,z=5,i;
i=x<y<z;
printf("%d",i);

Answer 1

x<y<z
Think it as (x<y)<z
x<y is a condition which always leads to either true or false.
True means 1
False means 0
So, (x<y) => 10<20 => true => 1
Now, (x<y)<z => 1<5 => true =>1

So, i=1

Answer 2

it will show a warning
the answer is for only x<y after that it will give you a warning

when we compare elements
the answer is given as true or false

in programing true = 1

and false = 0

here x < y is true ( so it is true = 1 )

it means x < y becomes 1

then 1 < z which is also true = 1 ( it will throw warning )

so i becomes 1;

hence final answer is 1.

/****************/

if you compare only x<y it will give you no warning but
compairing x<y<z will throw warnings

Answer 3

I can see HOW i is equal to 1
Your assignment expression results as true for 10 is less than 20.

The WHY is a bit beyond my pay grade.... Perhaps someone else will have an explanation on why this behaves like that.

If you just want to assign i with the smallest value of your other 3 variables, I think a repeated ternary might be the way to do it. Again, this isn't something I'm 100% on the syntax, but maybe;

i = x ? x<y : y ? y<z : z;

Answer 4

Ok thnx guys i understod it