68. Predict the output or error(s) of the following c code

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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);
}

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.