is it because of garbage value?

Question

https://code.sololearn.com/caN0CU5hYxBY/#c

Answer 1

The C standard does not define what order the arguments will be evaluated. So when one of the arguments reassigns a value that is used in another argument, then the result may vary. When I ran it, the result (0 50 0) implied that the arguments were evaluated right-to-left. A different order of evaluation would give a different result. Do not write code like this!

Answer 2

No its not garbage .1 50 1 will be the Output of your code

Answer 3

Jayakrishna🇮🇳 but am confused i think Output should be 1 50 1 here Values will store in stack formate so first x==35 this will store means left to right .

Answer 4

😅🐍🐍😆
There happening first k>40, k=50, k==35, right to left....
so output 0 50 0
In the stack it store tokens by "top to bottom" but evaluation is depends on precedence of Operators.
Generally expression evaluation is left to right (in equal precedence) , but
In a function call in C, the order in which the argument expressions are processed is not defined. And printf is a function also, in which is sequenced first for evaluation is depends on compiler specifications....
So different compiler result different values....
In c/c++, you can encounter this undefined behaviour...
Check out these....
May help....
https://en.m.wikipedia.org/wiki/Sequence_point
https://gateoverflow.in/62411/undefined-behaviour-in-c

Answer 5

Your code has sequence point issue.. You may get different result in different compilers.. depending on left to right, or right to left evaluation.
So it may 0 50 0 or 1 50 1
Not garbage value.