Perameter passing

#1

in swap function, we are sending variables simply and it looks it is being passed by reference how. Is it following a pass by reference by default?

#2

Yes. The swap function follows a pass by reference as the swapped values are reflected in the main program.

#3

but how it is implementing this functionality…it is not same with the normal function?

#4

void swap(int &a, int &b)
{
int temp = a;
a = b;
b = temp;
}

int main()
{
int a=5, b=10;
cout<<a<<" “<<b<<endl;
swap(a,b);
cout<<a<<” "<<b<<endl;
}

Output of this code will be -
5 10
10 5

#5

@gauravlodhi21 Please mark your doubt as resolved if you are satisfied.

#6

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