main()
{
int a = 65;
printf(“%d %o %x”,a,a,a);
}
Output
65 101 41
Please explain me.How it is coming like that?
Answers were Sorted based on User's Feedback
Answer / neha
It prints the value of in decimal, octal, hexadecimal
format respectively.
| Is This Answer Correct ? | 6 Yes | 0 No |
Answer / ruchika thakur
in octal,divide 65 by 8 and collect the remainder.it will
return 101....and in hexadecimal,divide 65 by 16 and again
collect the remainder...you will get 41......
THANKS
| Is This Answer Correct ? | 5 Yes | 2 No |
Answer / chandan
printf(�%d %o %x�,a,a,a);
1) We can use ? sign instead of " sing in pintf statement .
2)First o/p value 65 ,is the decimal value of int a.
3)2nd o/p value 101 , is the octal value of int a.
i.e base is 8.
8^2 8^1 8^0
1 0 1
it Works 8^2 *1 + 8^1 *0 + 8^0*1 = 64*1 + 8*0 + 1*1=64+0+1=65
it is actual input decimal value.
Similarly,
4)3rd o/p value 41 , is the Hexadecimal value of int a.
i.e base is 16.
16^1 16^0
4 1
it Works 16^1 *4 + 16^0*1 = 16*4 + 1*1=64+1=65
it is actual input decimal value.
| Is This Answer Correct ? | 2 Yes | 0 No |
can any one tel me wt is the question pattern for NIC exam
Given a number N, product(N) is the product of the digits of N. We can then form a sequence N, product(N), product(product(N))… For example, using 99, we get the sequence 99, 99 = 81, 81 = 8. Input Format: A single integer N Output Format: A single integer which is the number of steps after which a single digit number occurs in the sequence. Sample Test Cases: Input #00: 99 Output #00: 2 Explanation: Step - 1 : 9 * 9 = 81 Step - 2 : 8 * 1 = 8 There are 2 steps to get to this single digit number. Input #01: 1137638147
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