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main()
{
int i=5;
printf("%d%d%d%d",i++,i--,i);
}
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Answer / 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 |
some garbage value , 4,5,5...
why in this o/p garbage value is because only 3 parameters
are passed but we have assigned 4 control strings , where
one control string is useless, so for that compiler will
print garbage value....
| Is This Answer Correct ? | 2 Yes | 4 No |
Answer / sravankumar
printf() function evaluates from right to left
printf("\n %d %d %d",i++,i--,i);
4 5 5
<- <- <- <- <-evaluation of expression
but prints as the way we mentioned in printf() function
i.e first i = 5
then i--= 5 because it is post decrement
then i++= 4 this because i is decremented in above, and
not incremented immediately because is post
increment
So output is : 4 5 5
| Is This Answer Correct ? | 0 Yes | 3 No |
Answer / sunil5a2
4 5 5
printf excutes form lefthand side onwords..
| Is This Answer Correct ? | 0 Yes | 5 No |
#define PRINT(int) printf("int = %d ",int) main() {< BR> intx,y,z; x=03;y=02;z=01; PRINT(x^x); z<<=3;PRINT(x); y>>=3;PRINT(y); }
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