main()
{
int i = 10;
printf(" %d %d %d \n", ++i, i++, ++i);
}
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Answer / dhivya
11 11 13
++i which means i is first incremented and then printed thus i value becomes 11,
i++ which means i is printed and then it is incremented, since the value of i is now 11, it is printed and then incremented to 12.
again ++i , increments i to 13 and then prints.
| Is This Answer Correct ? | 60 Yes | 34 No |
Answer / dr. sanyasi naidu pasala
13 11 11
First the value of i was assigned to right most variable ++i. In this first i value which is 10 will be incremented to 11, then assigned to i and printed as 11. Then that 11 is passed to the middle variable which is i++. In this first the value 11 is assigned to i, print the value of i as 11 and then incremented to 12. This 12 is now passed to left most variable ++i. In this the value 12 is first incremented to 13 then assigned to i and print as 13. Even though generally the evaluation will be taken place from right most variable to left most variable, the evaluation process may vary from operating system to operating system.
| Is This Answer Correct ? | 24 Yes | 8 No |
Answer / krityangan
The Answer is 13 11 13
because the post increment will printed first and then pre.
i=10
the ++i=11,i++=12,++i=13,but in c the compiler will print ++p which is now 13 and after that when the compiler come to i++ it will print 11 because in the pre addition the previous vale is printed first andthan i= 12.
| Is This Answer Correct ? | 21 Yes | 10 No |
Answer / pooja alagarsamy
when compiled as a program, it gives this output:
13 11 13
| Is This Answer Correct ? | 1 Yes | 0 No |
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