C++:用 const 成员替换类数组的元素

Question

我正在尝试修改具有 const 成员的对象数组:

enum Bar {
    Baz,
    Qux,
    Quux
};

class Foo {
    public:
    Foo(Bar a, int b): a_(a), b_(b) {};

    private:
    const Bar a_;
    const int b_;
};

int main(int argc, char* argv[]) {
    Bar b[] = {
        Baz,
        Baz
    };

    // This works fine
    b[0] = Qux;

    Foo f[] = {
        Foo(Baz,42),
        Foo(Qux,32)
    };

    // This doesn't
    f[0] = Foo(Quux,3);

    return 0;
}

但是编译器不允许我:

$ make test
g++     test.cc   -o test
test.cc: In member function ‘Foo& Foo::operator=(const Foo&)’:
test.cc:7:7: error: non-static const member ‘const Bar Foo::a_’, can’t use default assignment operator
test.cc:7:7: error: non-static const member ‘const int Foo::b_’, can’t use default assignment operator
test.cc: In function ‘int main(int, char**)’:
test.cc:31:22: note: synthesised method ‘Foo& Foo::operator=(const Foo&)’ first required here 
make: *** [test] Error 1

我确信编译器有它的原因，我很想知道为什么代码不能正常工作。

我也想知道如何对 f 数组进行预期的更改。

现在，下面的代码为我完成了工作，但它看起来很不对:

#include <cstring>
#include <iostream>

enum Bar {
    Baz,
    Qux,
    Quux
};

class Foo {
    public:
    Foo(Bar a, int b): a_(a), b_(b) {};

    /*Foo &operator=(Foo const& f) {
     return f;
    }*/

    const Bar a_;
    const int b_;
};

int main(int argc, char* argv[]) {
    Bar b[] = {
        Baz,
        Baz
    };

    // This works fine
    b[0] = Qux;

    Foo f[] = {
        Foo(Baz,42),
        Foo(Qux,32)
    };

    // This doesn't
    //f[0] = Foo(Quux,3);

    // This does...
    Foo foo1(Quux, 344);
    memcpy(&f[0], &foo1, sizeof(foo1));

    std::cout << "Hi " << f[0].b_ <<"\n";
    return 0;
}

我希望有一个不涉及 memcpy 但仍以所需方式更改数组的解决方案。

Answer 1

数组与它无关。

Foo a, b;
a = b;

也应该无法编译，因为编译器不知道如何分配给 const 的值。