UGC NET Computer Science July 2018 - II | Question 5

Question 5

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5. Given below are three implementations of the swap( ) function in C++ :

(a)	(b)	(c)
void swap(int a, int b) {    int temp;    temp = a;    a = b;    b = temp; } int main() {   int p = 0, q = 1;   swap(p, q); }	void swap(int &a, int &b) {    int temp;    temp = a;    a = b;    b = temp; } int main() {  int p = 0, q = 1;  swap (p, q); }	void swap(int *a, int *b) {    int * temp;    temp = a;    a = b;    b = temp; } int main() {  int p = 0, q = 1;  swap (&p, &q); }

Which of these would actually swap the contents of the two integer variables p and q ?

1. (1) (a) only
2. (2) (b) only
3. (3) (c) only
4. (4) (b) and (c) only

Answer : (2) (b) only

Explanation Question 5

• Implementation (a) performs the swap on the local variables. So, the original variables are not changed.
• Implementation (b) swaps the the memory location of a and b (to which they are pointing to) inside swap(), hence p and q also get swapped in main(). Swap() function is accepting the alias of the variable (int &a) - it will evaluate parameter as int &a=p;. To swap values implementation (b) uses pass by reference without pointers.
• Implementation (c) is incorrect because inside swap function definition there is incorrect assignment of values at the addresses of a and b.

Hence, implementation (b) reflects the swapped value to the passed variable.

So, option (2) is right answer.

Function implementation to swap values using pass by reference without pointer:
void swap(int &n1, int &n2) {
    int temp;
    temp = n1;
    n1 = n2;
    n2 = temp;
}

Function implementation to swap values with pass by reference using pointers:
void swap(int* n1, int* n2) {
    int temp;
    temp = *n1;
    *n1 = *n2;
    *n2 = temp;
}

Reference : Example 1: Passing by reference without pointers

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