返回指针的方法，指向另一个标头中声明的数组对象，
我遇到了两个相互交织的问题。
首先，我想要一个指向堆上对象的指针数组。 （在另一个标头中声明的对象）
其次，我想让一个方法返回指向其中一个对象的指针。
我当前的代码是一些摸索的结果，并且会失败，因为我不能在没有完全声明它的情况下使用“bar”作为返回类型。但我看不出还有什么办法可以解决这个问题。我尝试使“getBar”成为指向函数的指针，但随后我不知道如何使其访问 **barArray 而不是成员方法。
任何帮助将不胜感激：D
我的代码：
foo.h
#ifndef FOO_H
#define FOO_H
//forward declaration
class bar;
class foo
{
public:
//constructor
foo(int x);
//method
bar * getBar(int y);
private:
int howManyBars;
bar **barArray;
};
#endif
foo.cpp
#include "foo.h"
#include "bar.h"
//constructor
foo::foo(int x)
{
howManyBars = x;
barArray = new bar *[howManyBars];
for (int i=0; i < howManyBars ; i++)
{
barArray[i] = NULL; //set all pointers to NULL
}
}
//method
bar * foo::getBar(int y)
{
y = (y - 1);
// if the pointer is null, make an object and return that
if (barArray[y] == NULL)
{
barArray[y] = new bar();
}
return barArray[y];
}
bar.h
#ifndef BAR_H
#define BAR_H
#include <iostream>
class bar
{
public:
void test(){std::cout << "I'm alive!\n";};
};
#endif
I'm getting a little stuck with two entwined problems.
First, I want to have an array of pointers to objects on the heap. (objects that are declared in another header)
Second, then I want to have a method return a pointer to one of those objects.
My current code is the result of a bit of fumbling, and will fail because I can't use "bar" as a return type without fully declaring it. But I can't see how else to solve the problem. I tried to make "getBar" a pointer to a function, but then I don't know how to make it access **barArray without it being a member method.
Any help would be much appreciated :D
My code:
foo.h
#ifndef FOO_H
#define FOO_H
//forward declaration
class bar;
class foo
{
public:
//constructor
foo(int x);
//method
bar * getBar(int y);
private:
int howManyBars;
bar **barArray;
};
#endif
foo.cpp
#include "foo.h"
#include "bar.h"
//constructor
foo::foo(int x)
{
howManyBars = x;
barArray = new bar *[howManyBars];
for (int i=0; i < howManyBars ; i++)
{
barArray[i] = NULL; //set all pointers to NULL
}
}
//method
bar * foo::getBar(int y)
{
y = (y - 1);
// if the pointer is null, make an object and return that
if (barArray[y] == NULL)
{
barArray[y] = new bar();
}
return barArray[y];
}
bar.h
#ifndef BAR_H
#define BAR_H
#include <iostream>
class bar
{
public:
void test(){std::cout << "I'm alive!\n";};
};
#endif
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评论（1）
除了一些拼写错误之外，这编译得很好：
bar * foo:getBar(int y)应该是：
3.
应该是：
Other than a few typos, this compiles fine:
bar * foo:getBar(int y)should be:
3.
should be: