C Programming - Structures, Unions, Enums - Discussion

Because enum values are initialised from zero.

Know that enum values are initialising from 0 only how it would get 2 and 1

Hi Ranjith,

Kindly read the program very carefully.

The order of values assigned are

stud1 = pass; // here pass = 0
stud2 = absent; // here absent = 2
stud3 = fail; // here fail = 1

So,

printf("%d %d %d\n", stud1, stud2, stud3);

produces the output as 0, 2, 1.

Hope you understand. Have a nice day!

How it assigning the value 0 for pass 2 for absent and 1 for fail. Wats the logic behind this? Please tell me in detail.

Hi Priya,
look at to the program carefully:
notice in line no. 3(where enum has declared) the order of declaration is pass=0,fail=1,then absent=2. So the order of declaration of enum is important,after that wherever we will use these variable the order will remain same it doen't matter in which order we have written it.
int main()
{
enum status {pass, fail, absent};
enum status stud1, stud2, stud3;
stud1 = pass;
stud2 = absent;
stud3 = fail;
printf("%d %d %d\n", stud1, stud2, stud3);
return 0;
}

Thanks xyz.

I satisfied with your ans.

Thank you friends your explanations helped me to get it clear.

Thats good explanation XYZ.

You should have write your name anyhow I am fully satisfied with that.

Good !

Hai, If you are declare any variables in enum the default initalization is started from 0 and sequentially they are increasing. So they didn't give any initaliztion in our programm that's why bydefault initaliztion will be takes place.

So pass=0, fail=1, abscent=2.