Question

class data
{
private:
    int ID;
    string address,name;
public:
    data(int i,string a,string n):ID(i),address(a),name(n){}
    friend class SetData;
};
class SetData
{
private:
    data obj(43,"185 Awan Market","Talha"); //this is where the error happens

public:
    void show()
    {
        cout<<"Name is: "<<obj.name<<endl;
        cout<<"Address is: "<<obj.address<<endl;
        cout<<"ID is: "<<obj.ID;
    }
};

Answer 1

<强> C ++ 03

它属于构造函数的mem-initializer：

include_once('../db.php');

session_start();

$seller_name = $_SESSION['seller_name'];
echo $seller_name;

  if (!empty($_POST['flag_image'])){
    $flags_array = $_POST['flag_image'];
    $flags_string = implode(",",$flags_array);
  }else{
     $flags_string = '';
  }

$cont_eu = '';

   // Assign Flags to Continents
    foreach ($flags_array as $flag_image){  
     $qry_find_flag_continent = " SELECT continent FROM flags WHERE flag_image = '$flag_image' ";
     $result = $dbdetails->query($qry_find_flag_continent);

 while($row = mysqli_fetch_array($result)) {
    $continent = $row["continent"];


    if ($continent == 'EU') {
     $cont_eu .= $flag_image;
    }   
} 
echo $cont_eu;

<强> C ++ 11

您必须使用大括号或相同的初始化程序。

class SetData
{
private:
    data obj;

public:
    SetData() : obj(43,"185 Awan Market","Talha")
    {
    }
    // Rest goes here...
};

为什么不允许使用括号，请参阅Non-static data member initializers提案。向下滚动到＆＃34; Kona提出的关于标识符范围的问题＆＃34;

类范围查找的动机是我们希望能够把任何东西都放在我们可以的非静态数据成员的初始化器中放入mem-initializer而不会显着改变语义（模数直接初始化与复制初始化）：
// Fine
data obj{43,"185 Awan Market","Talha"};
// Fine, too
data obj = data(43,"185 Awan Market","Talha"); //this is where the error happens
不幸的是，这会使“（表达式列表）”的初始化者在解析声明时表单不明确：
int x();

struct S {
    int i;
    S() : i(x()) {} // currently well-formed, uses S::x()
    // ...
    static int x();
};

struct T {
    int i = x(); // should use T::x(), ::x() would be a surprise
    // ...
    static int x();
};
一种可能的解决方案是依赖现有规则，如果a 声明可以是一个对象或一个函数，然后它是一个函数：
struct S {
    int i(x); // data member with initializer
    // ...
    static int x;
};

struct T {
    int i(x); // member function declaration
    // ...
    typedef int x;
};
类似的解决方案是目前应用另一个现有规则仅在模板中使用，如果T可以是类型或其他类型，那就是别的了;如果我们真的是这个意思，我们可以使用“typename” 一种类型：基本上
struct S {
    int i(j); // ill-formed...parsed as a member function,
              // type j looked up but not found
    // ...
    static int j;
};
这两种解决方案都会引入可能存在的微妙之处   被许多用户误解（正如许多问题所证明的那样）   comp.lang.c ++关于为什么“int i（）;”在块作用域中没有声明a   default-initialized int）。

本文提出的解决方案是仅允许初始化器   “= initializer-clause”和“{initializer-list}”形式。那   在大多数情况下解决歧义问题，例如：
struct S {
    int i(x); // unabmiguously a data member
    int j(typename y); // unabmiguously a member function
};

Answer 2

不允许以这种方式初始化非静态数据成员。您应该使用支撑或相等的初始化程序

class SetData
{
private:
    // data obj = {43,"185 Awan Market","Talha"}; is also valid
    data obj{43,"185 Awan Market","Talha"};

（CFR）。 non-static data members initialization

替代解决方案：构造函数初始化列表

class SetData
{
private:
    data obj;

public:
    SetData() : obj(43,"185 Awan Market","Talha") {}

    void show() 
    ...
};

至于为什么非静态数据成员初始化不支持括号，我建议阅读这篇文章：Why C++11 in-class initializer cannot use parentheses?

Answer 3

您无法像这样初始化内联对象，您必须在构造函数初始值设定项列表中执行此操作：

class SetData
{
private:
    data obj;

public:
    SetData() : obj(43,"185 Awan Market","Talha") {}
    ...
};

C ++表示数字常量之前的预期标识符

3 个答案: