main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
- 9 Answers
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Answer / susie
Answer :
10
Explanation:
The variable i is a block level variable and the visibility
is inside that block only. But the lifetime of i is lifetime
of the function so it lives upto the exit of main function.
Since the i is still allocated space, *j prints the value
stored in i since j points i.
| Is This Answer Correct ? | 78 Yes | 5 No |
Answer / vishu
the answer is that
int i varibale is part of int*j block code ,but outside the
block of code i variable also show their existanse.if we
write a code after the int*j block of code .
int*h
{
h=&i
}
printf("%d",*h);
}
| Is This Answer Correct ? | 11 Yes | 3 No |
Answer / bipin chandra sai.s
actually j has beeen assigned the addresss of i so the ans
will be the value present in the address location 10
| Is This Answer Correct ? | 7 Yes | 1 No |
Answer / jerome.s,final year eee,adhipa
There i-is initialised by 10.
and j-also initialised by address of i.
so *j is the value in the address of j.
therefore,
*j=i=10.
OUTPUT:
10
| Is This Answer Correct ? | 7 Yes | 1 No |
Answer / ashish p
The answer is undefined.
int *j;
{ //prolog
int i=10;
j = &i;
}//epilog
in the above code , at the prolog level the variables are
pushed into a un-named function space on the stack. Whereas
at epilog level the variable i dies.
J contains address of valid memory location but invalid
contents. Since i's memory is release back, any other
program can claim it and over-ride the contenets. Unless
then if we try to print the content using J it will give us
the value 10.
Which is not recommended it is something like returning
reference to the local variable in a function.
| Is This Answer Correct ? | 1 Yes | 1 No |
Answer / govind verma
i think ans will be 10 because here is the concept of dagling pointer......
| Is This Answer Correct ? | 0 Yes | 0 No |
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