But why would you have both functions in one and the same class definition? Having both allows you to: call the function on a mutable object, and modify the result if you like; and call the function on a const object, and only look at the result. With only the first, you couldn’t call it … Read more
Yes, you can use them together. And yes, it should exactly match the declaration in the translation unit it’s actually declared in. Unless of course you are participating in the Underhanded C Programming Contest 🙂 The usual pattern is: file.h: extern const int a_global_var; file.c: #include “file.h” const int a_global_var = /* some const expression … Read more
const, in Rust, is short for constant and is related to compile-time evaluation. It shows up: when declaring constants: const FOO: usize = 3; when declaring compile-time evaluable functions: const fn foo() -> &’static str These kinds of values can be used as generic parameters: [u8; FOO]. For now this is limited to array size, … Read more
consts are compile time constants. Meaning that their value has to be assigned during compile time, unlike vals, where it can be done at runtime. This means, that consts can never be assigned to a function or any class constructor, but only to a String or primitive. For example: const val foo = complexFunctionCall() //Not … Read more
You are not assigning to the const variable. You are instead assigning to the function parameter that you gave the same name. That function parameter is a non-const copy of the variable so you are allowed to assign to it. I will try to collect all my comments into a more fully explained answer. In … Read more
const is a compile-time feature. It doesn’t prevent you from shooting yourself in the foot; that’s what the warning is for.
Normally a temporary object (such as one returned by a function call) has a lifetime that extends to the end of the “enclosing expression”. However, a temporary bound to a reference generally has it’s lifetime ‘promoted’ to the lifetime of the reference (which may or may not be the lifetime of the calling function), but … Read more
int & geta(void) const { return *a; } // good? int & getb(void) const { return b; } // obviously bad In a const-function, every data member becomes const in such way that it cannot be modified. int becomes const int, int * becomes int * const, and so on. Since the type of a … Read more
It’s NOT portable to do so. std::pair requirements are laid out in clause 20.3. Clause 17.5.2.3 clarifies that Clauses 18 through 30 and Annex D do not specify the representation of classes, and intentionally omit specification of class members. An implementation may define static or non-static class members, or both, as needed to implement the … Read more
Because you’re thwarting the purpose of const, which is to keep you from modifying the argument. So if you cast away the constness of something, it’s pointless and bloating your code, and it lets you break promises that you made to the user of the function that you won’t modify the argument. In addition, using … Read more