If((!0 && 2) || (!1)) EXPLANATION(C)

Question

Can somebody explain this question .tq
If((!0 && 2) || (!1)){
printf("true");
}
Else{
Print("false");
}

Answer 1

it turns into a bool condition.

you can read it as:
if ((!false && true) || !true)

which is same as writing:
if ((true && true) || false)

which turns into:
if (true || false)

you think you understand?

Answer 2

It’s a boolean conditions
It’s like as
If((!false && true) || (!true))
Then

If((true&&true) || (false))
Finally
If(true || false)

Answer 3

true

Answer 4

is it true??

Answer 5

!0 is equal to 1
in C any non-zero value is true so (1 && 2) both values are true, which means first condition is true.

Your use of logical OR in if-statement means only one of two conditions need to be true for true to be printed. And the first one is.

Answer 6

Thanks everyone for helping

Answer 7

In if condition
If (!0&&2)||!1)
Here !0 is true so one condition is true here and u used or if one condition will be true its will give true so answer will be print true

Answer 8

Which part of these is confusing you if I may ask?

Answer 9

The whole if statement...😂I don't even understand how !0 && 2 will have an answer

Answer 10

Yup

Answer 11

True