Show the output from the following program. Assume the address of \"a\" is 0012F

Question

Show the output from the following program. Assume the address of "a" is 0012FF20, "b" is 0012FF40, "c" is 0012FF60, and "d" is 0012FF80.

#include <iostream>
using namespace std;

struct abc
{
    int ax;
    abc *ay;
};

void main()
{
    abc a = { 10, NULL }, b = { 20, &a }, c = { 40, &b }, d = { 80, &c };

    abc *e = &d;

    cout << e->ay->ay->ay << endl;
}

Show the output from the following program. Assume the address of "a" is 0012FF20, "b" is 0012FF40, "c" is 0012FF60, and "d" is 0012FF80.

#include <iostream>
using namespace std;

struct abc
{
    int ax;
    abc *ay;
};

void main()
{
    abc a = { 10, NULL }, b = { 20, &a }, c = { 40, &b }, d = { 80, &c };

    abc *e = &d;

    cout << d.ay->ay << endl;
}

Show the output from the following program. Assume the address of "a[0]" is 0012FF20, "a[1]" is 0012FF40, "a[2]" is 0012FF60, and "first" is 0012FF80.

#include <iostream>
using namespace std;

struct quiz
{
    int n1;
    int n2;
    quiz *n3;
};
void main()
{
    quiz a[3] = { { 1, 2, &a[1] }, { 3, 4, &a[2] }, { 5, 6, &a[0] } };
    quiz *first = &a[2];

    cout << first->n3->n3 << endl;

} //main

Explanation / Answer

1)Output is 0012FF20

Because e points to d , d points to c, c points to b and b points to a
when we do e->ay->ay->ay we get a's address so its 0012FF20


2) Output is 0012FF40

Because e points to d , d points to c, c points to b and b points to a
when we do d.ay->ay we get b's address so its 0012FF40

2) Output is 0012FF40

Because a[0] has a[1] address , a[1] has a[2] address , a[2] has a[0] address
first->n3 gives us a[0] and a[0]->n3 gives us a[1]
Thus we get a[1] address so its 0012FF40


Thanks

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