Why Can't We Overload Functions Based on the `const` Qualifier for Non-Pointer, Non-Reference Types?

Understanding Functions with const Arguments and Overloading

In programming, overloading allows multiple functions with the same name but different parameters to coexist in the same class. However, when attempting to overload based on the constness of a non-pointer, non-reference type, a compiler error arises. This article explores the reason behind this restriction and alternative approaches.

Consider the following code fragment:

class Test {
public:
    Test() {}
    int foo(const int) const;
    int foo(int);
};

When compiling this code, the compiler will issue an error indicating that the function foo cannot be overloaded. This is because the constness of a non-pointer, non-reference type does not affect the function signature. In the example above, both functions have the same signature: int foo(int).

To understand why this restriction exists, consider the following situation:

Test obj;
int variable = 0;
obj.foo(3);  // Calls the const function
obj.foo(variable);  // Intends to call the non-const function

If overloading were allowed based on constness, the compiler would be unable to determine which function to call when obj.foo(variable) is invoked. This is because, when passing by value, the value is copied and the const designation is irrelevant to the function call.

Therefore, overloading based on the constness of a non-pointer, non-reference type is not permitted in C . As an alternative, one could create two separate functions with different names, for example:

class Test {
public:
    Test() {}
    int fooConst(const int);
    int fooNonConst(int);
};

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