英文:
C++ function call in cout leads to "Reference to non-static member function must be called"
问题
以下是翻译好的部分:
我想直接将一些属性打印到 cout。幸运的是,我找到了一种针对静态调用的解决方案,就像这样:
// 某处:
ostream & Foo(ostream &ostr){
ostr << "Foo";
return ostr;
}
还有另一个静态调用,可以像这样直接调用:
// 其他地方:
void doSomething(){
cout << 123 << Foo << endl;
}
这很好地工作,并且简单地打印出 "123Foo"。但是,我想将其用于我拥有的对象,以便打印其属性。在我的类 A 中,我有一个类 B 的属性 b。在那里,函数的实现类似于之前的静态函数:
ostream & Foo(ostream &ostr) {
ostr << "my foo is 123";
return ostr;
}
但是它无法编译。我还尝试指定命名空间,例如:
ostream & B::Foo(ostream &ostr) { ... }
然后假设我可以像这样直接调用对象的函数:
cout << 123 << b->Foo << endl;
但它总是告诉我有一个错误:"对非静态成员函数的引用必须调用"。
我就是无法搞定这个。我做错了什么?这是否可能?
注意:我不想重载 << 运算符或其他任何内容。我不想要一个 "toString()" 方法或类似的东西,但我想了解为什么这个实例方法不能像其他地方一样被调用。作为一个 Java 开发者,我本来以为像 "My object " + b.foo() 这样的东西是可能的。但似乎是不行的。
英文:
I wanted to print some properties directly to cout. Luckily I found a solution for static calls like this:
// somewhere:
ostream & Foo(ostream &ostr){
ostr << "Foo";
return ostr;
}
and another static call where it can be called directly like that:
// somoewhere else:
void doSomething(){
cout << 123 << Foo << endl;
}
This works fine and prints simply "123Foo". But I want to use this for an object I have, so I could print its properties. In my class A I have a property b of the class B. There the function is implemented similar to the static one before:
ostream & Foo(ostream &ostr) {
ostr << "my foo is 123";
return ostr;
}
but it doesn't compile. I also tried to specify the namespace like
ostream & B::Foo(ostream &ostr) { ... }
and assumed I could just call the object's function like this:
cout << 123 << b->Foo << endl;
but it always tells me there is an error: "Reference to non-static member function must be called".
I just can't get this done. What am I doing wrong? Is this even possible?
NOTE: I don't want to overload the << operator or anything. I don't want a "toString()" method or anything similar but would like to understand why this instance method can not be called like anywhere else.
As a Java developer I would have guessed something like "My object " + b.foo() would be possible. But it seems, it is not.
答案1
得分: 0
问题在于你的独立函数 foo
是一个函数,而 b->foo
在C++中不是一个函数。换句话说,后者绑定了它的 this
参数,所以除了函数的行为之外,它还需要一些额外的数据。你可以使用显式lambda函数使其成为有效的 std::function
。
cout << 123 << [&b_object](ostream& out) { return b_object->foo(out); } << endl;
如果这是你对这个函数唯一的预期用法,那么你可以直接从 B::Foo
返回它。
std::function<ostream&(ostream&)> B::foo() {
return [&this](ostream& out) { out << "my foo is " << my_foo; return out; };
}
然后你可以这样写:
cout << 123 << b_object->foo() << endl;
但是,正如评论中所指出的,更常见的做法是使用专门的数据类型。如果你要打印的属性很简单,那么只需从 foo()
返回它或者它的引用即可。如果不是,或者如果你希望进行一些复杂的自定义打印操作,我建议编写自己的自定义类,并封装这个自定义打印行为。
class ViewOfObjectB {
private:
B& owner;
public:
ViewOfObjectB(B& owner) : owner(owner) {}
friend ostream& operator<<(ostream&, const ViewOfObjectB&);
};
ostream& operator<<(ostream& out, ViewOfObjectB& obj) {
// 你想要打印的内容放在这里...
return out;
}
class B {
public:
ViewOfObjectB foo() {
return ViewOfObjectB(&this);
}
};
与使用lambda函数的思路相同,但由于有大量的隐藏复杂性,我们使用了一个具有命名的视图类,可以在代码增长并变得更加复杂时独立地记录这个复杂性。你仍然可以这样调用它:
cout << 123 << b_object->foo() << endl;
所有这些调用都是静态的(在"没有虚函数表涉及"的意义上,而不是 static
关键字的意义上),所以一个聪明的优化器应该能够为你将所有内容内联。
英文:
The problem is that your standalone foo
is a function, whereas b->foo
is, according to C++, not one. Namely, the latter binds its this
argument, so it requires some extra data on top of just the function's behavior. You can make it a valid std::function
with an explicit lambda.
cout << 123 << [&b_object](ostream& out) { return b_object->foo(out); } << endl;
and if that's your only intended use case for this function, then you can return it from B::Foo
directly.
// Be careful with lifetimes! The return value cannot
// outlive `this`!
std::function<ostream&(ostream&)> B::foo() {
return [&this](ostream& out) { out << "my foo is " << my_foo; return out; };
}
Then you can write
cout << 123 << b_object->foo() << endl;
But, as indicated in the comments, it's much more common to use a specialized datatype for this. If the property you're trying to print is cheap, then just return it or a reference to it from foo()
and you're done. If not, or if you want some complicated custom printing operation to happen, I recommend writing your own custom class and encapsulating this custom print behavior.
class ViewOfObjectB {
private:
B& owner;
public:
ViewOfObjectB(B& owner) : owner(owner) {}
friend ostream& operator<<(ostream&, const ViewOfObjectB&);
};
ostream& operator<<(ostream& out, ViewOfObjectB& obj) {
// Whatever you want to print goes here ...
return out;
}
class B {
public:
ViewOfObjectB foo() {
return ViewOfObjectB(&this);
}
};
Same idea as using a lambda, but since there's a ton of hidden complexity, we make that complexity explicit with a named view class that we can document independently as our code grows and becomes more complicated. You still call it using
cout << 123 << b_object->foo() << endl;
All of these calls are static (in the sense of "no vtable involved", not in the sense of the static
keyword), so a clever optimizer should be able to inline it all for you.
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