Answer:
Explanation:
b) Each stage requires one clock cycle; ... Calculate how many clock cycles will take execution of this segment on the regular (non- pipelined) ... Instruction. Clock cycle number . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 ... It is possible ... It means that all stages of 5-stage pipeline are always busy (no stalls) during the task.
Answer:
#include <bits/stdc++.h>
using namespace std;
bool isPalindrome(string str)
{
char a,b;
int length = str.length();
for (int i = 0; i < length / 2; i++)
{
a=tolower(str[i]);//Converting both first characters to lowercase..
b=tolower(str[length-1-i]);
if (b != a )
return false;
}
return true;
}
int main() {
string t1;
cin>>t1;
if(isPalindrome(t1))
cout<<"The string is Palindrome"<<endl;
else
cout<<"The string is not Palindrome"<<endl;
return 0;
}
Output:-
Enter the string
madam
The string is Palindrome
Enter the string
abba
The string is Palindrome
Enter the string
22
The string is Palindrome
Enter the string
67876
The string is Palindrome
Enter the string
444244
The string is not Palindrome
Explanation:
To ignore the cases of uppercase and lower case i have converted every character to lowercase then checking each character.You can convert to uppercase also that will also work.
The answer is (a.) DOS
DOS stands for Disk Operating System. MS-DOS is the operating system which has a command interface similar to the command prompt of Windows 7 and Windows 8. Among the options, DOS is the only OS which is has a command line interface. Others have graphics.
Answer:
The advantage for the above condition is as follows:-
Explanation:
- If a user creates a defined constant variable and assigns a value on its and then uses that variable instead of the value, then it will a great advantage.
- It is because when there is a needs to change the value of that variable, then it can be done when the user changes the value in one place. There is no needs to change the vale in multiple places.
- But if there is a value in multiple places instead of a variable and there is no constant variable, then the user needs to change the value in multiple places.
Answer
First part:
The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.
Second part:
The invalid bit sequence are option a. 01001000 and d. 11100111
Explanation:
Explanation for first part:
In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.
If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.
If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.
Explanation for second part:
A valid odd parity bit sequence will always have odd number of 1s.
Since in option a and d, total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.
And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.
