我想将具有相同基类的多个类存储在std::vector
. 经过一些研究,我发现我必须使用指针来防止对象切片。但是,当我创建向量、向其添加元素并返回它时,生成的向量没有正确的值。
例如,这是我的两个类:
class Base {
public:
int var0;
}
class Derived : public Base {
public:
int var1;
}
这是一个简单的print
功能。作为一项规则,应该有的所有实例,以及Base
应该有的所有var0 == 23
实例。Derived
var0 != 23
void print(Base& value) {
if (value.var0 == 23) {
std::cout << "Base: " << value.var0 << std::endl;
} else {
Derived d = (Derived&) value;
std::cout << "Derived: " << d.var0 << ", " d.var1 << std::endl;
}
}
首先,这确实像我想要的那样工作:
int main() {
int num = 10;
std::vector<Base*> vec;
for (int i = 0; i < num; i++) {
if (i % 2 == 0) {
Base b;
b.var0 = 23;
vec.push_back(&b);
} else {
Derived d;
d.var0 = 17;
d.var1 = 42;
vec.push_back(&d);
}
}
// ....
for (int i = 0; i < num; i++) {
print(*vec.at(i));
}
}
这打印:
Base: 23
Derived: 17,42
Base: 23
Derived: 17,42
Base: 23
Derived: 17,42
Base: 23
Derived: 17,42
Base: 23
Derived: 17,42
现在,我希望这个向量由一个函数返回,所以我创建了一个函数:
std::vector<Base*> createVector(int num) {
std::vector<Base*> vec;
for (int i = 0; i < num; i++) {
if (i % 2 == 0) {
Base b;
b.var0 = 23;
vec.push_back(&b);
} else {
Derived d;
d.var0 = 17;
d.var1 = 42;
vec.push_back(&d);
}
}
return vec;
}
int main() {
int num = 10;
std::vector<Base*> vec = createVector(num);
// ....
for (int i = 0; i < num; i++) {
print(*vec.at(i));
}
}
这打印:
Derived: 2293232,0
Derived: 17,42
Derived: 2293232,0
Derived: 17,42
Derived: 2293232,0
Derived: 17,42
Derived: 2293232,0
Derived: 17,42
Derived: 2293232,0
Derived: 17,42
这不是我想要的。我希望它像其他功能一样打印。
有没有办法解决这个问题?有什么办法可以让整个派生类的事情做得更好一点吗?