Why this program is printing 1?

Question

https://code.sololearn.com/c3nOT8fB9w7w/?ref=app

Answer 1

int x=10, y=20, z=5, i;
i=x<y<z
=
i=(x<y)<z;
👇
1. 10<20 = true, (true = 1);
2. 1<5 = true;
3. i = true, i = 1;

Answer 2

Tanay Bhatt but bro if y<z is false then it should print 0.
Because 0 is false and 1 is true.
I think so...

Answer 3

Given condition is true
In C program for boolean conditions
If true it will return the value as 1
Else false it will return 0
/*This will vary on different languages*/

So if you print the i in %d formatter it shows 1

Answer 4

The precedence of the operators and associativity plays a role here

Answer 5

Vasiliy Thank you very much😄

Answer 6

x<y = true = 1: 10< 20
1<5=true=1
i=1
😎

Answer 7

x< y = 10 < 20 ?
Yes , true = 1

1 < 5 ?
Yes , i = true = 1

Answer 8

Bcz it a condition so check and ans is true so 1

Answer 9

You are comparing nos.
int x =10,y=20,z=5,i
i=(x<y)<z //(10<20) it is true and returning 1, in the next part 1<5//result of (x<y)<z =1
so that's why you got 1

Answer 10

It is a condition
and 1 is true

Answer 11

Theres no bracket in x<y<z, and what Tanay Bhatt is saying is absolutely correct.. but the compiler reads value from left to right so x<y is computed first then the resultant with z so output is 1(also operation is same so no precedence issue)..

Answer 12

sorry not tanay, Yamin Mansuri .. Am new to this app so mistake .. gomenasai