void main()
{
int i=1;
printf("%d%d%d",i,++i,i++);
}
Cau u say the output....?
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Answer / subash
221
because it starts executing from right to left
so it first executes i++(the increment will happen when
entire printf statement is executed so now ++i which means
it is the pre incremant so now i becomes 2
| Is This Answer Correct ? | 2 Yes | 3 No |
Answer / kiran
The output will be 1 2 2...
First the assigned value of i(1) is printed. Then (++i) prints the Incremented value of i which is 2 will be printed.
Now as the Postfix operator only prints thhe value first and then increments...so the value of i is again 2..
| Is This Answer Correct ? | 4 Yes | 5 No |
Answer / we r the new inventors
the explanation above are also correct but think they are somewhat wrong because it was post increment of i.so value of i will become 2..
| Is This Answer Correct ? | 1 Yes | 2 No |
Answer / we are the new inventors
above answers are also right but i think they are somewhat wrong..because its a post increment of i so ans will become 2...
| Is This Answer Correct ? | 0 Yes | 1 No |
Answer / sums
The ans is obv 3 3 1...
The explanations given are also correct
For all those u r giving other answers without even giving the reason should at least run the program in "gcc" before writing anything
| Is This Answer Correct ? | 2 Yes | 4 No |
Answer / pratik
answer is 1 1 2 ..
it is an example of stack type ..
| Is This Answer Correct ? | 0 Yes | 4 No |
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