返回一个指向数组的指针
假设您有一个带有指向数组的私有指针的类。如何使用 getter 来访问（或有效复制数据），以便可以在不同的变量中访问它。
class MyClass
{
private:
double *x;
public:
myClass();
virtual ~MyClass();
double* getX() const;
void setX(double* input);
};
MyClass::MyClass()
{
double foo[2];
double * xInput;
foo[0] = 1;
foo[1] = 2;
xInput = foo;
setX(xInput);
}
void MyClass::setX(double * input)
{
x = input;
}
double * MyClass::getX() const;
{
return x;
}
int main()
{
MyClass spam(); // Construct object
double * bar = spam.getX(); // This doesn't work
}
在这种情况下，bar[0]和bar[1]等于乱码：-9.2559631349317831e+061。
Suppose you have a class with a private pointer to an array. How can you use a getter to access (or effectively copy the data) so you can access it in a different variable.
class MyClass
{
private:
double *x;
public:
myClass();
virtual ~MyClass();
double* getX() const;
void setX(double* input);
};
MyClass::MyClass()
{
double foo[2];
double * xInput;
foo[0] = 1;
foo[1] = 2;
xInput = foo;
setX(xInput);
}
void MyClass::setX(double * input)
{
x = input;
}
double * MyClass::getX() const;
{
return x;
}
int main()
{
MyClass spam(); // Construct object
double * bar = spam.getX(); // This doesn't work
}
In this case, bar[0] and bar[1] are equal to jibberish: -9.2559631349317831e+061.
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评论（3）
它不会构造一个对象，该对象声明一个名为
spam的函数，该函数不接受任何参数并返回MyClass。此默认构造一个对象：有关更多信息，请谷歌最令人烦恼的解析。
更新：正如@Mark Ransom 指出的那样，您的代码还存在另一个问题。在构造函数中创建一个数组，然后设置 x 指向该数组。然而，一旦构造函数完成执行，数组的生命周期就结束了，因此进一步访问
x将会崩溃（如果你足够幸运）。That does not construct an object, that declares a function called
spamthat takes no arguments and returns aMyClass. This default constructs an object:For more information google the most vexing parse.
Update: As @Mark Ransom pointed out, there is another problem with your code. In your constructor you create an array, and then set
xto point to such array. However the array lifetime ends once the constructor finishes execution, so further access toxwould crash (if you are lucky enough).只是猜测，程序崩溃或显示错误的输出。这是因为构造函数正在设置一个指向本地数组的指针，该指针会离开作用域并在构造函数末尾被销毁。
Just a guess, the program is crashing or showing the wrong output. This is because the constructor is setting a pointer to a local array, which leaves scope and gets destroyed at the end of the constructor.
那可能应该是 *bar 而不是 bar[0]。
That should probably be *bar instead of bar[0].