Can C References Be Reassigned, or Is Something Else Happening?

Can References Be Reassigned in C ?

In C , references are often touted as immutable entities that must be initialized once and for all. However, a recent code snippet has raised questions about this principle.

Code Snippet:

int i = 5, j = 9;
int &ri = i;
cout << "ri is : " << ri << "\n";
i = 10;
cout << "ri is : " << ri << "\n";
ri = j;
cout << "ri is : " << ri << "\n";

Observations:

• The code compiles successfully without errors.
• The output shows that ri initially points to i and then changes to refer to j.

Question:

Does this code truly reassign the reference ri or is something else happening?

Answer:

No, ri is still a reference to i. The apparent reassignment is actually a modification of i through the reference ri.

Explanation:

When a reference is declared (e.g., int &ri = i), it binds to the object (i in this case) and remains linked to it throughout the program. The code ri = j does not reassign ri but rather modifies the value of i through the reference ri.

To prove this, one can print the addresses of ri and i using &ri and &i, which would show they remain the same. Additionally, if ri were reassigned to j, it would no longer be possible to modify i through ri, which is not the case in the given code.

Conclusion:

While references appear to be reassignable in the code snippet, they are not. Instead, they indirectly modify the object to which they refer. Const references (e.g., const int &cri = i) prevent such modifications and enforce true immutability.

The above is the detailed content of Can C References Be Reassigned, or Is Something Else Happening?. For more information, please follow other related articles on the PHP Chinese website!