Given the following program what will the output be? Please explain, not just gi

Question

Given the following program what will the output be? Please explain, not just give me the answer. I know it is 11 31

#include using

namespace std;

void one(int x, int& y);

void two(int& s, int t);

int main()

{

int u = 1;

int v = 2;

one(u, v);

cout << u << " " << v << endl;

two(u, v);

cout << u << " " << v << endl;

return 0;

}

void one(int x, int& y)

{

int a;

a = x;
x = y;

y = a;

}

void two(int& s, int t)

{ int b;

b = s - t;

s = t + b + 2;

t = 4 * b;

}

Explanation / Answer

ANSWER: I have run your code and it is giving the same output as you mention in this question.and explanation I added in the code as a comment you can check.

CODE:

#include <iostream>
using namespace std;

void one(int x, int& y);

void two(int& s, int t);

int main()

{

int u = 1;

int v = 2;

one(u, v);// Here you are passing 1 and 2 in this function after going int his function you are passing the address of v and in this function swapping is performed so in this funtion v is changed to its address and after this function v contain 1 but u are passing as a value so it contains its original value 1 that's why it is printing 1 1.

cout << u << " " << v << endl;

two(u, v);// Here you are passing u and v value again and in this function you are passing the address of u and changes its value to 3 in the funtion so thats why it is printing 3 1 aftwer this function same as above concept is used here

cout << u << " " << v << endl;

return 0;

}

void one(int x, int& y)

{

int a;

a = x;
x = y;

y = a;
}

void two(int& s, int t)

{ int b;

b = s - t;// here b will be 1-1 =0

s = t + b + 2;// s will be 1+0+2=3 and it is changes its address so it is permanent change.

t = 4 * b;

}

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