Answer:
couple.py
def couple(s1,s2):
newlist = []
for i in range(len(s1)):
newlist.append([s1[i],s2[i]])
return newlist
s1=[1,2,3]
s2=[4,5,6]
print(couple(s1,s2))
enum.py
def couple(s1,s2):
newlist = []
for i in range(len(s1)):
newlist.append([s1[i],s2[i]])
return newlist
def enumerate(s,start=0):
number_Array=[ i for i in range(start,start+len(s))]
return couple(number_Array,s)
s=[6,1,'a']
print(enumerate(s))
print(enumerate('five',5))
Explanation:
Answer:
The explanation to this question is given below in the explanation section.
Explanation:
#include <iostream>
#include <string>
using namespace std;
int main()
{
int score;
cout << "Enter Score: \n";
cin>>score;
if (score == 100)
//whether the value of the score variable is equal to 100
{
cout << "Perfect ";
//If so, it should print “Perfect”
}
else
{
cout << "Nice Try ";
}
}
Answer:
b) Bounded Waiting
Explanation:
int currentThread = 1;
bool thread1Access = true;
bool thread2Access = true;
thread1 { thread2 {
While (true) {
While (true)
{
while(thread2Access == true)
{
while(thread1Access == true)
{
If (currentThread == 2) {
If (currentThread == 1)
{
thread1Access = false; thread2Access = false;
While (currentThread == 2);
While (currentThread == 1);
thread1Access = true; thread2Access = true;
} }
/* start of critical section */ /* start of critical section */
currentThread = 2 currentThread = 1
… ...
/* end of critical section */ /* end of critical section */
thread1Access = false; thread2Access = false;
… ...
} }
} }
} }
It can be seen that in all the instances, both threads are programmed to share same resource at the same time, and hence this is the bounded waiting. For Mutual exclusion, two threads cannot share one resource at one time. They must share simultaneously. Also there should be no deadlock. For Progress each thread should have exclusive access to all the resources. Thus its definitely the not the Progress. And hence its Bounded waiting.
