在 c++ 中从 void* 转换为对象数组
我在让它工作时遇到问题，
class A {
public:
A(int n) {
a = n;
}
int getA() {
return a;
}
private:
int a;
};
int main(){
A* a[3];
A* b[3];
for (int i = 0; i < 3; ++i) {
a[i] = new A(i + 1);
}
void * pointer = a;
b = (A* [])pointer; // DOESNT WORK Apparently ISO C++ forbids casting to an array type ‘A* []’.
b = static_cast<A*[]>(pointer); // DOESN'T WORK invalid static_cast from type ‘void*’ to type ‘A* []’
return 0;
}
而且我无法使用泛型类型来满足我的需要。
提前致谢。
I'm having problems getting this to work,
class A {
public:
A(int n) {
a = n;
}
int getA() {
return a;
}
private:
int a;
};
int main(){
A* a[3];
A* b[3];
for (int i = 0; i < 3; ++i) {
a[i] = new A(i + 1);
}
void * pointer = a;
b = (A* [])pointer; // DOESNT WORK Apparently ISO C++ forbids casting to an array type ‘A* []’.
b = static_cast<A*[]>(pointer); // DOESN'T WORK invalid static_cast from type ‘void*’ to type ‘A* []’
return 0;
}
And i can't use generic types for what i need.
Thanks in advance.
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数组在 C 中是二等公民（因此在 C++ 中也是如此）。例如，您无法分配它们。并且很难将它们传递给函数而不降级为指向其第一个元素的指针。
在大多数情况下，指向数组第一个元素的指针可以像数组一样使用 - 除非您不能使用它来获取数组的大小。
当您编写时，
a会隐式转换为指向其第一个元素的指针，然后将其转换为void*。由此，您无法取回数组，但可以获取指向第一个元素的指针：（
注意：在指向不相关类型的指针之间进行转换需要
reinterpret_cast，除了转换为void*是隐式的，从void*到任何其他指针，这可以使用static_cast来完成。）Arrays are second-class citizen in C (and thus in C++). For example, you can't assign them. And it's hard to pass them to a function without them degrading to a pointer to their first element.
A pointer to an array's first element can for most purposes be used like the array - except you cannot use it to get the array's size.
When you write
ais implicitly converted to a pointer to its first element, and that is then casted tovoid*.From that, you cannot have the array back, but you can get the pointer to the first element:
(Note: casting between pointers to unrelated types requires a
reinterpret_cast, except for casts tovoid*which are implicit and fromvoid*to any other pointer, which can be done using astatic_cast.)也许你想做什么
？
static_cast版本也是可能的。Perhaps you mean to do
?
A
static_castversion is also possible.