Latest Miscellaneous Company Exam Aptitude Question SOLUTION: int a; printf("%u\n", (int)(&a+1)-(int)(&a)); it prints sizeof(a)=4; but i can't understand ,pls can any one explain this.

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(#M40035037) MISCELLANEOUS COMPAN Programming question c-language Keep an EYE

int a;
printf("%un", (int)(&a+1)-(int)(&a));
it prints sizeof(a)=4;
but i can't understand ,pls can any one explain this.

Asked In Miscellaneous Company Exam krishna (10 years ago)

Unsolved Read Solution (4)

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Saraswathy  (10 years ago)

sizeof(a) ll be 2 only...

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Saraswathy  (10 years ago)

Assume the address assigned to "a" is 2000... So it l take 2 bytes as its an "integer" variable... So "a" ll hold 2000 nd 2001...

printf("%un", (int)(&a+1)-(int)(&a));
The above line ll print "2"... Since &a=2000... When u increment an address variable, it ll increment as many bytes as its holding... So it ll increment 2 bytes [ ie 2002 nd thats is converted to int (int)] ... Next (int) (&a) = 2000
So 2002 - 2000 = 2...
Hence its printing 2...
If u print sizeof(a), again it ll print 2 only... Bcozzz sizeof integer variable is always "2" only...The increment of address never affects the sizeof the variable...

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Pallab Das  (10 years ago)

explain which is main thing .......

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Pallab Das  (10 years ago)

this output depends upon compiler in ANSI C it will print 4 .

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