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void main()
{
for(int i=0;i<5;i++);
printf("%d",i);
}
What is the output?..
- 32 Answers
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Answer / sudeshna
it is a declaration error.
since we declared variable i in the for-loop
and the for-loop is terminated by a semi-colon
so it cannot be accessed outside the for-loop
| Is This Answer Correct ? | 0 Yes | 2 No |
Answer / suggest
it will report error....
bcoz, integer variable i have scope inside the for loop only....
we cant access it in printf....bcoz for loop have one semicolon
| Is This Answer Correct ? | 0 Yes | 2 No |
Answer / medo
It's 5...if this is in the condition.(case 1)
But if the condition i<=5,the output will be 6.(case 2)
So the hand trace for the cace 1:
Memory__ |_|_|_|___|
i =0 |1|2|3|4|(5)|
i++ =1 |2|3|4|5| - |
it will print 5.
_-_-_-_-_-_-_-_-_-_-_
In the case 2:
Memory__ |_|_|_|_|___|
i =0 |1|2|3|4|5|(6)|
i++ =1 |2|3|4|5|6| - |
it will print 6.
| Is This Answer Correct ? | 10 Yes | 13 No |
Answer / samir isakoski
If this is a regular c
you cannot put in for loop, non declared integer
it must by declared before the for loop
from 0 to 5
0
1
2
3
4
beacouse it's start from zerro
| Is This Answer Correct ? | 4 Yes | 7 No |
We can't declare a variable in any part of the program
rather than the declaration part.
It doesn't matter whether you use a loop to print or not.
When the statements which are to be executed begins(Here the
looping statement)no declaration is possible in C.
You can do it in C++,C#,java etc.
| Is This Answer Correct ? | 6 Yes | 15 No |
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