使用参数调用类构造函数 - 请求“x”中的非类类型成员 [英] Calling class constructor with parameter - request of a member in 'x' which is of non-class type
问题描述
我有一个接受类B作为构造函数参数的类A. B类可以用int值构造。我原来的代码非常复杂，但我希望我已经将它简化为基本情况：
class B {
public：
explicit B（int a）：val（a）{}
private：
int val;
};
$ b $ class A {
public：
A（const B& val）：value（val）{};
void print（）{
//不做任何事
}
private：
B值;
};
int main（）{
int someTimeVar = 22;
A a（B（someTimeVar））;
.print（）;
}
这是我得到的错误代码：
$ g ++ test.cpp -Wall -O0
test.cpp：在函数'int main（）'中：
test .cpp：22：7：error：请求'a'中的成员'print'，它是非类类型'A（B）'
a.print（）;
^
test.cpp：20：9：warning：未使用的变量'someTimeVar'[-Wunused-variable]
int someTimeVar = 22;
$
我使用GCC（4.9.2 20150304（预发布）），platform：arch
对主函数进行的以下修改可以很好地编译：
int main（）{
A a（B（22））;
.print（）;
}
我很清楚使用A a（）;声明一个函数，而不是一个对象。但我没想到A a（B（some_val））也会这样做，并且我认为这是发生在这里的事情。
你有什么想法这是怎么回事？
编辑：感谢所有的答案，看起来我需要更多地研究最令人头疼的解析思路。 p>
顺便说一句，使用clang编译我的代码会提供更有用的错误信息和解决方案：
$ clang test.cpp
test.cpp：21：8：warning：圆括号被解析为函数声明[-Wvexing-parse]
A a（B（someTimeVar）） ;
^ ~~~~~~~~~~~~~~~
test.cpp：21：9：注意：添加一对圆括号来声明变量
A a（ B（someTimeVar））;
（）
test.cpp：22：6：error：成员引用基类型'A（B）'不是结构体或联合体
a.print（） ;
〜^ ~~~~~
1个警告和1个错误产生。
这个问题在stackoverflow上有自己的标记。最令人烦恼的解析
维基百科对问题及其解决方案进行了清晰的描述。 https://en.wikipedia.org/wiki/Most_vexing_parse 。
行
TimeKeeper time_keeper（Timer（））;
可以被解释为：
- TimeKeeper类的变量time_keeper的变量定义，传递类Timer的匿名实例或
- 函数time_keeper的函数声明，返回TimeKeeper类型的对象，有一个单一的（未命名的）参数
是一个返回类型Timer的函数（并且没有输入）。 （请参阅
函数对象＃在C和C ++中）
大多数程序员都期待第一个，但C ++标准要求它
被解释为第二个。
解决方法是将括号添加到参数中：
A a（（B（22）））;
或者像其他人指出的那样使用通用初始化，如
A a {B {22}};
I have a class A that accepts class B as a constructor parameter. Class B can be constructed from int value. My original code is quite complex, but I hope I've reduced it to the very base case:
class B {
public:
explicit B(int a) : val(a) {}
private:
int val;
};
class A {
public:
A(const B & val) : value(val) {};
void print() {
//does nothing
}
private:
B value;
};
int main() {
int someTimeVar = 22;
A a(B(someTimeVar));
a.print();
}
And this is the error code I'm getting:
$ g++ test.cpp -Wall -O0
test.cpp: In function ‘int main()’:
test.cpp:22:7: error: request for member ‘print’ in ‘a’, which is of non-class type ‘A(B)’
a.print();
^
test.cpp:20:9: warning: unused variable ‘someTimeVar’ [-Wunused-variable]
int someTimeVar = 22;
^
I use GCC (4.9.2 20150304 (prerelease)), platform: arch linux.
The following modification to the main function compiles fine:
int main() {
A a(B(22));
a.print();
}
I'm well aware that using A a(); declares a function, not a object. But I didn't expect that A a(B(some_val)) will do the same, and in my opinion this is what's happening here.
Do you have ideas why this is happening?
Edit: Thank you for all the answers, looks like I need to research more on most vexing parse idea.
BTW it turns out that compiling my code using clang provides more useful error message plus a solution:
$ clang test.cpp
test.cpp:21:8: warning: parentheses were disambiguated as a function declaration [-Wvexing-parse]
A a(B(someTimeVar));
^~~~~~~~~~~~~~~~
test.cpp:21:9: note: add a pair of parentheses to declare a variable
A a(B(someTimeVar));
^
( )
test.cpp:22:6: error: member reference base type 'A (B)' is not a structure or union
a.print();
~^~~~~~
1 warning and 1 error generated.
This problem has its own tag here on stackoverflow. most-vexing-parse
Wikipedia har a clear description of the problem and its solution. https://en.wikipedia.org/wiki/Most_vexing_parse.
The line
TimeKeeper time_keeper(Timer());could be disambiguated either as
- a variable definition for variable time_keeper of class TimeKeeper, passed an anonymous instance of class Timer or
- a function declaration for a function time_keeper which returns an object of type TimeKeeper and has a single (unnamed) parameter which is a function returning type Timer (and taking no input). (See Function object#In C and C++)
Most programmers expect the first, but the C++ standard requires it to be interpreted as the second.
The solution is to add parenthesis to the argument like:
A a( (B(22)) );
or as others have noted use universal initialization like
A a { B{22} };
这篇关于使用参数调用类构造函数 - 请求“x”中的非类类型成员的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
