C Programming - Structures, Unions, Enums - Discussion

@Manu. It is compiler dependent.

#include<stdio.h>

int main()
{
enum status {pass, fail, absent};
enum status stud1, stud2, stud3;

printf("%d %d %d %d \n", absent,stud1,stud2, stud3);
return 0;
}

Output: 2 51 51 8

Please explain why stud1, stud2 = 51 stud3 =8;

i = 0 1 2.

enum status {pass, fail, absent};

enum status stud1, stud2, stud3;

stud1 = pass; ->0

stud2 = absent; ->2

stud3 = fail; ->1

So 0 2 1

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

Simply enum start index from 0 by default if no any value is associated.

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.

Good !

Thats good explanation XYZ.

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

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

Thanks xyz.

I satisfied with your ans.

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;
}