“const"有多少和哪些用途?在 C++ 中? [英] How many and which are the uses of "const" in C++?

问题描述

As a novice C++ programmer there are some constructs that look still very obscure to me, one of these is const. You can use it in so many places and with so many different effects that is nearly impossible for a beginner to come out alive. Will some C++ guru explain once forever the various uses and whether and/or why not to use them?

解决方案

Trying to collect some uses:

Binding some temporary to reference-to-const, to lengthen its lifetime. The reference can be a base - and the destructor of it doesn't need to be virtual - the right destructor is still called:

ScopeGuard const& guard = MakeGuard(&cleanUpFunction);

Explanation, using code:

struct ScopeGuard { 
    ~ScopeGuard() { } // not virtual
};

template<typename T> struct Derived : ScopeGuard { 
    T t; 
    Derived(T t):t(t) { }
    ~Derived() {
        t(); // call function
    }
};

template<typename T> Derived<T> MakeGuard(T t) { return Derived<T>(t); }

This trick is used in Alexandrescu's ScopeGuard utility class. Once the temporary goes out of scope, the destructor of Derived is called correctly. The above code misses some small details, but that's the big deal with it.

---

Use const to tell others methods won't change the logical state of this object.

struct SmartPtr {
    int getCopies() const { return mCopiesMade; }
};

---

Use const for copy-on-write classes, to make the compiler help you to decide when and when not you need to copy.

struct MyString {
    char * getData() { /* copy: caller might write */ return mData; }
    char const* getData() const { return mData; }
};

Explanation: You might want to share data when you copy something as long as the data of the originally and the copie'd object remain the same. Once one of the object changes data, you however need now two versions: One for the original, and one for the copy. That is, you copy on a write to either object, so that they now both have their own version.

Using code:

int main() {
    string const a = "1234";
    string const b = a;
    // outputs the same address for COW strings
    cout << (void*)&a[0] << ", " << (void*)&b[0];
}

The above snippet prints the same address on my GCC, because the used C++ library implements a copy-on-write std::string. Both strings, even though they are distinct objects, share the same memory for their string data. Making b non-const will prefer the non-const version of the operator[] and GCC will create a copy of the backing memory buffer, because we could change it and it must not affect the data of a!

int main() {
    string const a = "1234";
    string b = a;
    // outputs different addresses!
    cout << (void*)&a[0] << ", " << (void*)&b[0];
}

---

For the copy-constructor to make copies from const objects and temporaries:

struct MyClass {
    MyClass(MyClass const& that) { /* make copy of that */ }
};

---

For making constants that trivially can't change

double const PI = 3.1415;

---

For passing arbitrary objects by reference instead of by value - to prevent possibly expensive or impossible by-value passing

void PrintIt(Object const& obj) {
    // ...
}

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