Question

C++ program. Please explain how the code resulted in the output. The code and output is listed below.

Code:

#include <iostream>
#include <string>

using namespace std;

int f(int& a, int b) {
   int tmp = a;
   a = b;

   if (tmp == 0) { cout << tmp << ' ' << a << ' ' << b << endl; }

   b = tmp;
   return b;
   return a;
}

int main() {
   int a = 0, b = 1, c = 8;

   f(b, c);
   cout << a << ' ' << b << ' ' << c << endl;

   a = f(a, b);
   cout << a << ' ' << b << ' ' << c << endl;

   return 0;
}

Output:

0 8 8

0 8 8

0 8 8

Answer 1

int &a is the reference variable, which means its an another variable for an already existing variable.

Function	Main
int f(int& a, int b) { 6. int tmp = a; 7. a = b; 8. if (tmp == 0) { cout << tmp << ' ' << a << ' ' << b << endl; } 9. b = tmp; 10. return b; 11. return a; }	int main() { 1. int a = 0, b = 1, c = 8; 2. f(b, c); 3. cout << a << ' ' << b << ' ' << c << endl; 4. a = f(a, b); 5. cout << a << ' ' << b << ' ' << c << endl;    return 0; }

variable a = 0, b = 1, c = 8

line 2: is calling function f(b,c)

line 6: tmp = 1 (a value is 1)

line 7 : a=8 (b is having 8 value) note : a is reference variable which is referencing b in main, so a value reflect in b.

line 8 : will not execute..(if condition is false)

line 9: b=1 (tmp is having 1 value)

line 10: return

line 3: print a , b , c [a is having 0, b is having 8(because of line 7) and c is having 8]

line 4: is calling a = f(a, b)

line 6: tmp = 0 (a value is 0)

line 7 : a=8 (b is having 8 value) note : a is reference variable which is referencing 'a' in main, so 'a' value reflect in 'a' of main.

line 8 : print tmp , a , b [tmp is having 0, a is having 8 and b is having 8 ](if condition is true)

line 9: b=0 (tmp is having 0 value)

line 10: return b

line 4: a = 0 (function returned 0 )

line 5: print a , b , c [a is having 0, b is having 8(updated by previous function call) and c is having 8]