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Old 08-24-2012, 02:42 AM   #1
tushar_pandey
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Union :: problem with declaration .


Code:
uinon Person
{
       int age_according_to_mother ;
       int age_according_to_father ;
}

int age_according_to_mother = 10 ;
int age_according_to_father = 20 ;

printf("Age_according_to_Mother :: %d" , age_according_to_mother );
the result is :: 20 ; why ....

& now where is our declaration .... means , we have defined age_according_to_mother = 10 ; age_according_to_father = 20 ; means they both got the space in memory , but we have used union so they are overlapped , but if they are overlapped than how our program finds or sets the value of age_according_to_mother = 20 !
 
Old 08-24-2012, 03:15 AM   #2
414N
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How can you even get said code to compile?
  • union is misspelled (uinon);
  • the union declaration is not semicolon-terminated
  • the two int variables don't even refer to the union fields
You should post a meaningful compilable portion of your code that's behaving strangely, as it seems you can compile it.
 
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Old 08-24-2012, 03:19 AM   #3
SIG_SEGV
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Put semicolon after the union definition..........
 
Old 08-24-2012, 03:29 AM   #4
tushar_pandey
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Code:
#include<stdio.h>

typedef union Person
{
        int age_according_to_mother ;
        int age_according_to_father ;
}person ;

int main()
{
        person u_1 ;

        u_1.age_according_to_mother = 10 ;
        u_1.age_according_to_father = 20 ;

        printf("%d",u_1.age_according_to_mother);

        printf("\n");
        return 0 ;
}

Last edited by tushar_pandey; 08-24-2012 at 03:32 AM.
 
Old 08-24-2012, 03:32 AM   #5
414N
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That's perfectly normal union behavior.
I guess you should make sure you got unions right...
 
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Old 08-24-2012, 03:34 AM   #6
tushar_pandey
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Quote:
Originally Posted by 414N View Post
That's perfectly normal union behavior.
I guess you should make sure you got unions right...
but can you please explain this ,

the result is :: 20 ; why ....

Quote:
& now where is our declaration .... means , we have defined age_according_to_mother = 10 ; age_according_to_father = 20 ; means they both got the space in memory , but we have used union so they are overlapped , but if they are overlapped than how our program finds or sets the value of age_according_to_mother = 20 ! how our program finds age_according_to_mother

Last edited by tushar_pandey; 08-24-2012 at 03:36 AM.
 
Old 08-24-2012, 03:40 AM   #7
SIG_SEGV
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So, then the answer is right here . I mean u_1.age1's value is printed as 20 here. whenever, you use the union, it is allocated with the memory space of its only one member(of largest size)
EX: union a{ int b; double c;};
Here union 'a' is allocated with 8bytes (Bcoz highest bytesized member is 'double' and is of 8bytes).

And a basic fact of union is that it can hold the value of a single member at any instant. And also all the members of union are given the same address (i.e beginning address of the union itself). I hope this is enough for you to crack the question
Convinced????????
 
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Old 08-24-2012, 03:43 AM   #8
tushar_pandey
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thanks ,

Quote:
And a basic fact of union is that it can hold the value of a single member at any instant. And also all the members of union are given the same address (i.e beginning address of the union itself). I hope this is enough for you to crack the question

it works ...
 
Old 08-24-2012, 03:53 AM   #9
414N
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If the issue is no more, mark the thread as solved, then.
 
Old 08-24-2012, 04:30 AM   #10
tushar_pandey
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414N , you seems like strict + dangerous guy !
 
Old 08-24-2012, 06:09 AM   #11
414N
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Strict? Maybe. I just suggested you to read again what unions are and how they work because your "issue" was, likely, a misunderstanding on the subject. I was going to explain a bit more, but SIG_SEV came first
Regarding the "dangerous" bit, I don't know how one can be dangerous by pointing someone else to read again its documentation...
 
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