指向類數(shù)據(jù)成員的指針“：*”我偶然發(fā)現(xiàn)了一個奇怪的代碼片段，它編譯得很好：class Car{
public:
int speed;};int main(){
int Car::*pSpeed = &Car::speed;
return 0;}為什么C+是否有指向類的非靜態(tài)數(shù)據(jù)成員的指針？什么這個奇怪的指針在實際代碼中使用嗎？
3 回答
嗶嗶one
TA貢獻1854條經(jīng)驗 獲得超8個贊
#include <iostream>class bowl {public:
int apples;
int oranges;};int count_fruit(bowl * begin, bowl * end, int bowl::*fruit){
int count = 0;
for (bowl * iterator = begin; iterator != end; ++ iterator)
count += iterator->*fruit;
return count;}int main(){
bowl bowls[2] = {
{ 1, 2 },
{ 3, 5 }
};
std::cout << "I have " << count_fruit(bowls, bowls + 2, & bowl::apples) << " apples\n";
std::cout << "I have " << count_fruit(bowls, bowls + 2, & bowl::oranges) << " oranges\n";
return 0;}
拉莫斯之舞
TA貢獻1820條經(jīng)驗 獲得超10個贊
// say this is some existing structure. And we want to use// a list. We can tell it that the next pointer// is apple::next.struct apple {
int data;
apple * next;};// simple example of a minimal intrusive list. Could specify the// member pointer as template argument too, if we wanted:
// template<typename E, E *E::*next_ptr>template<typename E>struct List {
List(E *E::*next_ptr):head(0), next_ptr(next_ptr) { }
void add(E &e) {
// access its next pointer by the member pointer
e.*next_ptr = head;
head = &e;
}
E * head;
E *E::*next_ptr;};int main() {
List<apple> lst(&apple::next);
apple a;
lst.add(a);}- 3 回答
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