问题描述
class fruit
{
public:
std::string apple;
std::string banana;
std::String orange;
};
class grocery
{
public:
std::vector<fruit> g_items;
std::string total_weight;
};
std::vector<grocery> shopping;
auto check_item = [&](std::string f_itm) -> void
{
for (std::size_t i = 0; i < shopping.g_items.size(); ++i)
{
std::cout << shopping.g_items.at(i).f_itm << std::endl;
}
};
check_item(apple);
check_item(banan);
check_item(orange)
如何调用 lambda 函数 check_item
,传递 fruit
对象的特定数据成员,因为它们都具有相同的类型 (std:string
)?
谢谢。
解决方法
试试这个:
#include <string>
#include <vector>
#include <iostream>
class fruit
{
public:
std::string apple;
std::string banana;
std::string orange; //std::string,not std::String
};
class grocery
{
public:
std::vector<fruit> g_items;
std::string total_weight;
};
std::vector<grocery> shopping;
auto check_item = [&](std::string(fruit::* f_itm)) -> void
{
for (std::size_t i = 0; i < shopping.size(); ++i) //you have two,vectors,so first iterate through shopping
{
for (std::size_t j = 0; j < shopping[i].g_items.size(); ++j) //then,iterate through g_items
std::cout << shopping[i].g_items[i].*f_itm << std::endl; //then,print
}
};
int main() {
check_item(&fruit::apple);
check_item(&fruit::banana); //you put banan,not banana
check_item(&fruit::orange); //remember the ;
}
您的原始代码有几个错误。然后,在您的 for 循环中,您有两个迭代两个向量,如注释所述。然后,它应该可以工作了。
,更改 lambda 以接收指向成员的指针,例如:
class fruit
{
public:
std::string apple;
std::string banana;
std::string orange;
};
class grocery
{
public:
std::vector<fruit> g_items;
std::string total_weight;
};
std::vector<grocery> shopping;
...
auto check_item = [&](std::string (fruit::*f_itm)) -> void
{
for (auto &g : shopping)
{
for (auto &f : g.g_items) {
std::cout << f.*f_itm << std::endl;
}
}
};
check_item(&fruit::apple);
check_item(&fruit::banana);
check_item(&fruit::orange);