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main()
{
float a=3.2e40;
printf("%d",a);
}
- 9 Answers
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- Satyam, I also Faced
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Answers were Sorted based on User's Feedback
Answer / pankaj bhalerao
i run this and i found that the answer is zero because a
is declared as float and it is print ans integer and its
mantissa part is also a float therefore the output is zero.
thank you...
| Is This Answer Correct ? | 12 Yes | 2 No |
Answer / pushpanjali panda
after running the program i got the answer is 0.
| Is This Answer Correct ? | 8 Yes | 0 No |
Answer / iftekhar
main()
{
int i=300*300/300;
printf("%d",i);
}
| Is This Answer Correct ? | 12 Yes | 9 No |
Answer / vignesh1988i
as for as i know ,
here however 'a' is a float variable when it comes
to printf , the floating numbers will be truncated and it
outputs as 3
thank u
| Is This Answer Correct ? | 11 Yes | 9 No |
Answer / deepali chandra
the o/p=0
because here we are trying to print the value in int form
which has been of float datatype.since float has a higher
precendence over int therefore it will give the o/p=0 and
similarly after this every output will be =0.this is the
function of the compiler that we cannot print the value of
a higher data type using a lower datatype.
| Is This Answer Correct ? | 0 Yes | 1 No |
Answer / siva
a is declared as float but it is printed as integer so it
must be wrong.........
| Is This Answer Correct ? | 0 Yes | 1 No |
it is not execute, bcz in the printf satement declared as
integer so answer is 0.
| Is This Answer Correct ? | 0 Yes | 1 No |
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