使用QVector时，没有对默认构造函数的匹配调用 [英] No matching call to default constructor, when using QVector

问题描述

我有一个B类，它创建A类的对象并调用该对象的方法。

I have a class B that creates an object of a class A and calls a method of the object.

ah

#ifndef A_H
#define A_H

class A
{
public:
    A(int);
    void function();
};

#endif // A_H

a.cpp

#include "a.h"

A::A(int x)
{

}
void A::function(){
    //Do something.
}

bh

#ifndef B_H
#define B_H
#include <QVector>
#include <a.h>


class B
{
public:
    B(int);
    QVector<A> list;
};

#endif // B_H

b.cpp

#include "b.h"

B::B(int y)
{
    list.append(A(y));
    list[0].function();
}  

问题是它不能编译。它返回没有匹配的函数调用'A：A（）'。我知道可以使用前向声明解决此问题，但这在这里不起作用，因为我想将函数称为函数。我也不想将整个A类都包含在B类中。

The problem is that this does not compile. It returns "no matching function to call 'A:A()'". I know that this can be solved with a forward declaration but this does not work here since I want to call the function "function". I also do not want to include the whole class A in the class B.

推荐答案

与许多Qt容器一样， QVector 的元素类型必须为您的版本中的可分配数据类型 。

As with many Qt containers, QVector's element type must be an assignable data type in your version.

与标准库不同， Qt将此定义为：

存储在各个容器中的值可以是任何可分配的数据类型。要获得资格，类型必须提供默认的构造函数，副本构造函数和赋值运算符。

The values stored in the various containers can be of any assignable data type. To qualify, a type must provide a default constructor, a copy constructor, and an assignment operator.

这真的很不幸，因为< a href = https://stackoverflow.com/q/33380402/560648>在您的示例中，实际上不需要默认的构造函数，实际上不需要 std :: vector 可以（合规地）使用没有元素的元素类型。

This is really unfortunate, because there's no practical need for a default constructor in your example, and indeed a std::vector would (compliantly) let you use an element type that doesn't have one.

QVector :: value（int ）函数确实依赖此属性，但您没有使用它！ Qt开发人员必须预先进行某种检查，而不是采用标准库的方法仅在实际需要时检查先决条件，否则这是代码的意外！

The QVector::value(int) function does rely on this property, but you're not using it! The Qt devs must be doing some kind of checks up-front, rather than taking the standard library's approach of "just check preconditions when they're actually needed", or else this is an "accident" of the code!

因此，直到更改此参数的5.13 ，您必须为 A 提供默认的构造函数，抱歉。

As a consequence, until 5.13 in which this was changed, you will have to give A a default constructor, sorry.

也不要忘记复制构造函数…以及对该 A :: function（）定义的适当限定。

Don't forget a copy constructor, too… and a proper qualification on that A::function() definition.

前向声明不能解决这个问题，您也不需要一个。实际上，在此特定程序中添加一个实际上不会执行任何操作。 ;）

A forward declaration will not solve this, neither do you need one. In fact, adding one to this particular program will do literally nothing. ;)

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