指针和引用作为const对象的成员变量 [英] Pointers and References as member variables of const objects

问题描述

以下代码可以正常编译。但是我不知道这是否合法的C ++。因此，更具体地说，如果我有一个const对象，是否可以通过该对象的指针/引用来修改变量？

The following code compiles fine. However I wonder if it is legal C++. So more specific, if I have a const object, am I allowed to modify variables through pointers/references of that object?

class Foo {
public:
    int* a;
    int& b;
    Foo(int* _a, int& _b) : a(_a), b(_b) {}
};

int main ( int argc, char* argv[] ) {
    int x = 7;
    const Foo bar(&x, x);

    *bar.a = 3; //Leagal?
    bar.b = 7; //Legal?
    return 0;
}

推荐答案

这是合法的，因为const-类的级别表示类成员是恒定的。 a 是一个指针，因此指针指向的地址是恒定的，但存储在该地址的值不必是恒定的。

It's legal, as const-ness of the class means that the class member is constant. a is a pointer, so the address the pointer points to is constant, but the value stored at that address need not be.

因此， bar.a 实际上是 int * const ，而不是 int const * 。

Hence bar.a is effectively an int * const, not an int const *.

由于初始化后，无论如何都不能引用另一个实体，对于 bar.b 是否将 bar 声明为 const 。

As, after initialization, a reference cannot be made to refer to another entity anyway, it does not matter for bar.b whether bar is declared const or not.

指针的常量变量是常量指针，而不是指向常量的指针。引用的常量变体是引用，而不是对常量的引用。

The constant variant of a pointer is a constant pointer, not a pointer to a constant. The constant variant of a reference is a reference, not a reference to a constant.

小题外话：无论如何，在与const-ness有关的成员中，您都应谨慎对待引用，因为以下内容可能会编译

Small digression: You should be careful with references as members anyway in connection with const-ness, as the following will probably compile

struct Y { int m_a; };
struct X {
   const Y & m_y;
   X (const Y & y) : m_y (y) { }
};

Y y;
y.m_a = 1;
X x (y); // or const X x (y) -- does not matter
// X.m_y.m_a == 1

y.m_a = 2;
// now X.m_y.m_a == 2, although X.m_y is supposed to be const

由于可以将指向非const的指针分配给指向const的指针，因此可以使用指针构建类似的示例。请记住， const 仅保证您不会通过此非常大的变量来修改变量，它不能保证完全不修改变量的内容。

As it is possible to assign a pointer to non-const to a pointer to const, you can build an analogous example with pointers. Remember that const does only guarantee that YOU will not modify a variable via this very variable, it cannot guarantee that the contents of the variable are not modified at all.

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