Description
int x;
int * ptr1, * ptr2; double * ptr3;
int x; int y; int * p = &x; p = &y; *p = 10; x = y + 20; p = &x; y = 25; *p = 50;
cout << *p << ” ” << x << ” ” << y << endl;
int *x = new int; int *y; *x = 60; y = x; *y = *y + *x; x = new int; *x = *y – 20; cout << *x << ” ” << *y << endl;
double *x = new double; double *y = new double; *x = 10; y = x; delete x; delete y; x = new double; *x = 20;
cout << *x << ” ” << *y << endl;
int * aPtr; // aPtr should point to array a int n;
int a[5] = { 1, 2, 3, 4, 5 };
// use pointer to get the first value of array // assign 2nd element of array to n
for (int i = 0; i <= 5; ++i) cout << aPtr[i] << endl;
++a;
int array[7] = { 4, 8, 9, 1, 13, 32, 20};
int * ptr = array; *ptr = *ptr + 5; ptr = ptr + 2;
*ptr = (*ptr) – *(ptr – 1); ptr++;
*ptr = 5 * (*ptr) – 2;
int info;
Node * next;
};
head current temp trail last Node *
Use the above list to answer questions 1 to 5.
temp = current; // Line 1
current = current->next; // Line 2
current->next = last; // Line 3
trail = current-> next; // Line 4
trail = trail->next; // Line 5
cout << current->info << ” ” << trail->info << endl; // Line 6
