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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 / vera
012345
not
0
1
2
3
4
cz their is no printf("\n");
first time i=0 it print 0
scnd time i=1 it print 1
last time i=4 it print 4
but not 5 cz when i=5 :i is not <5
so it will not print 5
10x
| Is This Answer Correct ? | 4 Yes | 5 No |
Answer / sanjeevi
Guys remember things
1) we cant declare loop variable inside for
2) here i is local to for loop alone..
| Is This Answer Correct ? | 1 Yes | 2 No |
Answer / chuimaya
It gives an error coz we cannot declare a variable inside the for loop.
| Is This Answer Correct ? | 0 Yes | 1 No |
Answer / ramveer
for the first loop i is 0 so 0 is printed
for 2nd time i++ i becomes 1 so 1 is printed
for 3rd time i becomes 2 so 2 gets printed
for 4th time i becomes 3 so it gets prited
for 5th time i becomes 4 so it gets printed
now for this time i has became 5 and this isn't in loop boudary as the condition is i<5 so its breaks out of loop.
so the ans is "0 1 2 3 4" .
| Is This Answer Correct ? | 9 Yes | 10 No |
Answer / erdem
Ofcourse
01234
(logic for start from 0 to 4 and no blank or tab)
| Is This Answer Correct ? | 1 Yes | 2 No |
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