Suppose that we have these two situations?
int p = 10;
int& q = p;
and
int p = 10;
int q = p;
Are not these two situations the same? I am little confused with the purpose of references, so please explain difference in these two.
4 Answers
int p = 10;
int& q = p;
In this case, q is for all practical purposes an alias for p. They share a memory location. If you modify q, you modify p. If you modify p, you modify q. q is just a different name for p.
int p = 10;
int q = p;
Here, q gets a copy of the value of p at the time when q is initialized. Afterwards, q and p are completely independent. Changing q does not affect p and changing p does not affect q.
In the second case, if you change the value of q it will not affect the value of p. In the first case, changing the value of q will also change the value of p and vice versa.
$ cat ref.cpp
#include <iostream>
int main () {
int p = 10;
int q = p;
int s = 10;
int& t = s;
q = 11;
t = 11;
std::cout << p << std::endl;
std::cout << q << std::endl;
std::cout << s << std::endl;
std::cout << t << std::endl;
s = 12;
std::cout << s << std::endl;
std::cout << t << std::endl;
}
$ g++ ref.cpp
$ ./a.out
10
11
11
11
12
12
$
Comments
In Second case p q are two independent integers. in first case both p q will point to same location in the memory. as you asked for the purpose of references. please go through call by value & call by reference. you can understand the use of references. go through the page. http://www.tutorialspoint.com/cplusplus/cpp_function_call_by_reference.htm
Comments
Changing a reference will change the underlying variable. This can be useful in situations like:
int count_odd = 0;
int count_even = 0;
int i;
....
// Create a reference to either count_odd or count_even.
int& count = (i%1) ? count_odd : count_even;
// Now update the right count;
count++;
This is rather artificial, but it becomes more useful with rather more complicated situations.
answered Feb 1, 2016 at 8:55
Martin Bonner supports Monica
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