Answer:
The first and last variable
Explanation:
Answer:
A. Technicians A and B
Explanation:
When we're talking about generic scanners and about all OBD-II codes, in this case, both technician A and B is the correct answer. Because we can scan all OBD-II codes with a generic scan.
But the technician A just says generic tools must be able to read all generic OBD-II codes and technicians B just says generic scan tools must be able to read manufacture OBD-II code, both are correct.
Answer:
the save command is located in the file menu
Answer:
#include <iostream>
#include <map>
using namespace std;
int main()
{
map<int, int> numbers;
cout << "Enter numbers, 0 to finish" << endl;
int number;
while (true) {
cin >> number;
if (number == 0) break;
numbers[number]++;
}
for (pair<int, int> element : numbers) {
std::cout << element.first << ": occurs " << element.second << " times" << std::endl;
}
}
Explanation:
One trick used here is not to keep track of the numbers themselves (since that is not a requirement), but start counting their occurrances right away. An STL map< > is a more suitable construct than a vector< >.
Answer:
In assembly language, two instructions control the use of the assembly language procedure.
CALL pushed the control to the return address onto the stack and transferred the control.
RET instruction returns the address that placed on the stack by a call instruction.
Explanation:
Action RET instruction
- The RET instruction pops the address and returns off the stack, which is pointed by the stack pointer.
- The stack is LIFO in memory at a particular location, and the pointer points offset from the stack location.
RET instruction does its job by consulting the register and memory state at the point when it is executed.
In RET instruction, only register and memory state is executed. Call instruction must save that address that figure out in a register and memory location.
