Answer:
answer:
#include <iostream>
#include<list>
using namespace std;
bool Greater(int x) { return x>3; } int main() { list<int>l; /*Declare the list of integers*/ l.push_back(5); l.push_back(6); /*Insert 5 and 6 at the end of list*/ l.push_front(1); l.push_front(2); /*Insert 1 and 2 in front of the list*/ list<int>::iterator it = l.begin(); advance(it, 2); l.insert(it, 4); /*Insert 4 at position 3*/ for(list<int>::iterator i = l.begin();i != l.end();i++) cout<< *i << " "; /*Display the list*/ cout<<endl; l.erase(it); /*Delete the element 4 inserted at position 3*/ for(list<int>::iterator i = l.begin();i != l.end();i++) cout<< *i << " "; /*Display the list*/ cout<<endl;
l.remove_if(Greater); for(list<int>::iterator i = l.begin();i != l.end();i++) cout<< *i << " ";
/*Display the list*/
cout<<endl; return 0;
}
Mechanical mouse has a ball that turns rollers inside. If friction is lost between the ball and the mousing surface, or between the ball and the rollers, the mouse fails to work. In order to assure good contact with the mousing surface, the ball must be fairly heavy. When you change directions with the mouse, you must make the ball change rolling directions--an action that inertia likes to prevent.
An optical mouse makes use of an LED and some optics to detect surface texture and the changes in it as the mouse is moved. There are no moving parts
Answer:
Sequence of popped values: h,s,f.
State of stack (from top to bottom): m, d
Explanation:
Assuming that stack is initially empty. Suppose that p contains the popped values. The state of the stack is where the top and bottom are pointing to in the stack. The top of the stack is that end of the stack where the new value is entered and existing values is removed. The sequence works as following:
push(d) -> enters d to the Stack
Stack:
d ->top
push(h) -> enters h to the Stack
Stack:
h ->top
d ->bottom
pop() -> removes h from the Stack:
Stack:
d ->top
p: Suppose p contains popped values so first popped value entered to p is h
p = h
push(f) -> enters f to the Stack
Stack:
f ->top
d ->bottom
push(s) -> enters s to the Stack
Stack:
s ->top
f
d ->bottom
pop() -> removes s from the Stack:
Stack:
f ->top
d -> bottom
p = h, s
pop() -> removes f from the Stack:
Stack:
d ->top
p = h, s, f
push(m) -> enters m to the Stack:
Stack:
m ->top
d ->bottom
So looking at p the sequence of popped values is:
h, s, f
the final state of the stack:
m, d
end that is the top of the stack:
m
Aerodynamic- Wind turbine, computational fluid dynamics, and Wind power
Hydrodynamics- Computational Fluid Dynamics, Hydraulics, and Microfluidics
