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int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
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Answer / susie
Answer :
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the
innermost block i is declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So
printf prints 30. In the next block, i has value 20 and so
printf prints 20. In the outermost block, i is declared as
extern, so no storage space is allocated for it. After
compilation is over the linker resolves it to global
variable i (since it is the only variable visible there). So
it prints i's value as 10.
| Is This Answer Correct ? | 6 Yes | 8 No |
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C Code 
