CODEWITHSUNDEEP
cconditional-statementslogical-operators
Arivarasan
Arivarasan

Reputation: 389

execution of conditional operators

int j=4;
(!j!=1)?printf("Welcome"):printf("Bye");

In the above code segment, according to me, first j!=1 will result in true and !true is false which must result in printing Bye but I get Welcome as the output.

Can anyone explain this one?

Upvotes: 0

Views: 83

Answers (5)

msc
msc

Reputation: 34608

! executed first because unary operator ! has a higher precedence than !=.

!4 become 0 then 0 != 1 become true.

So, output is Welcome.

Upvotes: 2

work.bin
work.bin

Reputation: 1108

The unary operator '!' has a higher precedence than '!='.

Read - https://www.tutorialspoint.com/cprogramming/c_operators_precedence.htm.

Upvotes: 2

GrahamS
GrahamS

Reputation: 10350

The Logical NOT operator ! has a higher precedence than the Not Equal To operator !=

So your condition is equivalent to ((!j) != 1)

See https://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B#Operator_precedence

Upvotes: 2

Charles
Charles

Reputation: 1396

This is because ! (NOT) has higher operator precedence than != so...

j = 4; // 4
!j // 0

In your condition, 0 != 1 will be true so "Welcome" is printed.

For your desired outcome, your condition would have to be !(j!=1).

Upvotes: 2

user1084944
user1084944

Reputation:

!j!=1 is (!j)!=1, not !(j!=1).

Upvotes: 2

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