#include <stdio.h> int main() { int a=8, b=0,c=0; if(!a<10&& !b||c) printf("Rahman"); elae printf("abdul"); return 0; }

Question

Explain
How it work

Answer 1

#include <stdio.h>

int main() {
int a=8, b=0,c=0;
if(!a<10&& !b||c)
printf("Rahman");
else printf("abdul");
return 0;
}

if(!a<10&& !b||c)
Is equals to
if( ((!a) <10) && (!b) || c)

!a = !8 = 0
!b = 1

So
if( (0<10) && 1 || 0)
Is equals to
if( true && 1 || 0)
Is equals to True

It should print "Rahman"

Answer 2

Abdul Rahman Khan firstly your code has some type and missing parenthesis the correct code should be like this
#include <stdio.h>

int main() {
int a=8, b=0,c=0;
if(!(a<10)&& !(b||c))
printf("Rahman");
else
printf("abdul");
return 0;
}
Now how "abdul " is printed? Because
8<10= true which is 1 in boolean the not operation is performed then so 1 become 0
0||0 gives 0 then not operation make it 1
Now 0 && 1 which evaluated as 0 because and(&&) is multiplication and or (||) is sum so it gives 0 and else part is evaluated and printed.

Answer 3

Precedence overrides left-to-right association Gordon

Answer 4

Both version of codes are answered above😊

Answer 5

Gordon && has higher precedence than ||
! operator has even more higher

Answer 6

https://code.sololearn.com/cVDb0edfnIgs/?ref=app

Answer 7

Gordon https://en.cppreference.com/w/cpp/language/operator_precedence

Answer 8

Thanks to all

Answer 9

If there is no parenthesis what will be answer

Answer 10

How

Answer 11

true && 1 || 0
Is
(true && 1) || 0
Or
true && (1||0)
?

Answer 12

That means
"a and b or c" is "(a and b) or c",
and
"a or b and c" is "a or (b and c)"?

Answer 13

Sequence of precedence, from high to low:
!
<
&&
||

Answer 14

true && 1 || 0 is equals to (true && 1) || 0

Answer 15

Thanks for the reference website.
I noticed the associativity of Left-to-right, then that explains the above code playground result.