Question

在一个非常简单的情况下，我有以下设置，其中，我只是想从类A的函数初始化一个常量静态成员（类foo）（单例和实例与此问题无关）：

class A
{
public:

    static A instance;
    A & getInstance() { return instance; }
    int i(){ return 10;}
    int j(){ return 20;}
};

class foo {
public:

    static const int ii = A::getInstance().i() * A::getInstance().j(); 
};
const int foo::ii;

int main()
{
    foo f;
    return 1;
}

目的是使用上述某些功能初始化成员ii。但它会产生以下错误：

$ c++ static_constant.cpp 
static_constant.cpp:14:30: error: ‘A::getInstance()’ cannot appear in a constant-expression
static_constant.cpp:14:42: error: a function call cannot appear in a constant-expression
static_constant.cpp:14:44: error: ‘.’ cannot appear in a constant-expression
static_constant.cpp:14:46: error: a function call cannot appear in a constant-expression
static_constant.cpp:14:53: error: ‘A::getInstance()’ cannot appear in a constant-expression
static_constant.cpp:14:65: error: a function call cannot appear in a constant-expression
static_constant.cpp:14:67: error: ‘.’ cannot appear in a constant-expression
static_constant.cpp:14:69: error: a function call cannot appear in a constant-expression

你可以帮帮我吗？会欣赏它。

Answer 1

您使用getInstance作为静态函数，但未声明static。

更改getInstance的声明：

static A & getInstance() { return instance; }

Answer 2

代码有几个问题，但这里有一个完整的可编辑样本：

class A
{
public:

    static A instance;
    static A & getInstance() { return instance; }
    int i(){ return 10;}
    int j(){ return 20;}
};

class foo {
public:

    static const int ii;
};
const int foo::ii = A::getInstance().i() * A::getInstance().j(); 
A A::instance;

如何使用函数初始化静态常量成员

2 个答案: