In the C Primer 5th Edition, the following distinction is made:
int f(int){ /* can write to parameter */} int f(const int){ /* cannot write to parameter */}
While these two functions appear indistinguishable, they indeed differ in their ability to update their parameters. However, they remain distinguishable in their function signatures.
The reason for this non-distinction lies in the "pass by value" nature of value-based parameters. When an object is passed to the function, a copy is created, which is the actual parameter modified within the function. The const qualifier at the top level does not affect the value of the copy, as the copy is not const const. Therefore, from the caller's perspective, both functions have the same effect.
Overloading of functions is based on the parameters provided by the caller. The constness of a parameter does not alter the functionality the called function provides, so it makes no logical sense to vary the implementation based on it. Consider the following code:
f(3); int x = 1 + 2; f(x);
The function f() is expected to behave identically in both cases, with or without the const qualifier. Providing different implementations could lead to confusion and errors.
In contrast to value-based parameters, references pass by reference to the actual object, not a copy. This allows for both overloading based on the constness of the reference and passing constness through function calls. For example:
const T& f(const F&); T& f(F&);
Here, the implementation can vary depending on whether the parameter is passed by const reference or non-const reference.
Despite the lack of distinction in function signatures, there is a way to emulate the desired behavior by using const references:
T f(F& x_ref) { F x = x_ref; // or const F if you won't modify it ...use x for safety... }
By passing the parameter as a const reference, the compiler will prohibit any modifications to the parameter. This ensures safety while providing a similar interface.
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