A leading 0 denotes octal notation, so the digits 8 and 9 are not used.
In normal decimal notation 12 means 1*10+2, in octal, it means 1*8+2.
The leading 0x denotes hexadecimal notation, here the digits are extended with 6 letters, a through f. 0x12 means 1*16+2.
Without any prefix, we have normal decimal notation.
Answer:
(a)
= 
Explanation:
To convert from binary to hexadecimal, convert each 4 binary digits to its hexadecimal equivalent according to the following;
<em>Binary => Hex</em>
0000 => 0
0001 => 1
0010 => 2
0011 => 3
0100 => 4
0101 => 5
0110 => 6
0111 => 7
1000 => 8
1001 => 9
1010 => A
1011 => B
1100 => C
1101 => D
1110 => E
1111 => F
(a) 1100 1111 0101 0111
=> Taking the first four binary digits : 1100
According to the table, the hexadecimal equivalent is C
=> Taking the second four binary digits : 1111
According to the table, the hexadecimal equivalent is F
=> Taking the third four binary digits : 0101
According to the table, the hexadecimal equivalent is 5
=> Taking the last four binary digits : 0111
According to the table, the hexadecimal equivalent is 7
Therefore, the hexadecimal representation of
1100 1111 0101 0111 is CF57
Answer: option A) First
In a circular linked list the last node points to the first node.
Explanation: As in a circular linked list, the node will point to its next node. Since the node next to last node is the first node hence last node will point to first node in a circular way. The elements points to each other in a circular path which forms a circular chain.
I think not but I'm not sure
Hello, since you did not specify a programming language, I wrote this algorithm in C++. Good luck!
<h2>
Code:</h2>
#include <iostream>
#include <vector>
std::vector<int> v;
int main(int argc, char* argv[]) {
while(1) {
int temp;
std::cout << "\nEnter a number: ";std::cin>>temp;
if(temp<0) {
std::cout << "\nEven number(s) is/are:\n---------------------\n";
for(int i=0;i<v.size();i++) {
if(v.at(i)%2==0) std::cout << v[i] << " ";
else continue;
}
std::cout << std::endl;
break;
}else {
v.push_back(temp);
}
}
return 0;
}