CODEWITHSUNDEEP
csyntax
user2578525
user2578525

Reputation: 191

Conditional expressions as arguments to printf

So i came across the following code in C Language

foo() {                     
    int v=10;
    printf("%d %d %d\n", v==10, v=25, v > 20);
}

and it returns 0 25 0 can anybody explain me how and why

Upvotes: 2

Views: 96

Answers (4)

Gopi
Gopi

Reputation: 19874

printf("%d %d %d\n", v==10, v=25, v > 20);

What you see is undefined behavior becuase the order of evalutaion within printf() is not defined.

The output can be explained as(Right to left evaluation)

v = 10 and hence v>20 is false so last `%d` prints 0
v = 25 Now v is 25 so second `printf()` prints out 25

Then you have

v ==10 which is false because v is 25 now. This is not a defined order of evaluation and might vary so this is UB

Upvotes: 3

Thiyagu
Thiyagu

Reputation: 17910

It is evaluated from right to left...

First it evaluates v > 20 its false so it prints 0

Next it sets v=25 an prints it

Next it check if v is 10. Its false so it prints 0 (the value of v is changed in the above step)

EDIT

This is the way your compiler evaluates it but the order of evalaution generally is undefined

Upvotes: 0

R Sahu
R Sahu

Reputation: 206697

Your code is subject to undefined behavior. Looks like in your platform,

v > 20 got evaluated first, followed by
v=25, followed by
v==10.

Which is perfectly standards compliant behavior.

Remember that those expressions could have been evaluated in any order and it would still be standards compliant behavior.

Upvotes: 0

kelsier
kelsier

Reputation: 4100

Your compiler seems to evaluate the function parameters from right to left.

So, v > 20 is evaluated first, then v=25 and then v==10.

Therefore you get the output 0 25 0

Upvotes: 0

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