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main()
{
int i=5;
printf("%d%d%d%d",i++,i--,i);
}
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Answer Posted / sravan kumer
Answer is 455848 in 'Turbo C++'
because here gave 4 %d's but given variables to print are 3.
So here Turbo C++ will evaluate first 3 parameters given as
---> first i will be evaluated so i=5 because printf() evaluates from right to left.
---> then i-- is 5 because it is post decrement so 1st prints value and then increments i=4
---> then i++ is post increments so 1st prints i value i.e 4
and then it will be incremented to 5.
---> so it printf will print 455 but there is another %d ,printf will handle those with a garbage values i.e 848 here.
so answer is 455848.i.e 455 is common after that some garbage value will be printed.
| Is This Answer Correct ? | 1 Yes | 0 No |
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