a09*_*935 6 c++ arrays string c++14
我正在使用C++ 14并尝试为每个循环创建一个打印出该数组的每个字符串的循环.我收到错误:
user.cpp:12:34:错误:从'char'类型的表达式初始化'std :: string&{aka std :: basic_string&}'类型的引用无效
for(std :: string&str:*(u-> favs)){
当我在foreach循环中将std :: string更改为auto时,它可以工作,但str成为favs数组中第一个字符串的单个字符.我的代码如下:
user.h
class User{
private:
public:
User(){
favs = new std::string[5]{std::string("Hello"), std::string("how"), std::string("are"), std::string("you"), std::string("?")};
}
~User(){
delete[] favs;
}
std::string lName;
std::string fName;
int age;
std::string *favs;
};
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user.cpp
#include <iostream>
#include "user.h"
void create(User *u){
std::cout << "Please enter first name: ";
std::cin >> u->fName;
std::cout << "Please enter last name: ";
std::cin >> u->lName;
std::cout << "Please enter age: ";
std::cin >> u->age;
for(std::string &str : *(u->favs)){
std::cout << str << std::endl;
}
std::cout << "\n";
}
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main.cpp中
#include <iostream>
#include <string>
#include "user.h"
int main(){
std::string command;
User user;
std::cout << "bp1: " << user.favs->size() << std::endl;
while(true){
std::cout << "Please enter a command (Create, Update, View, Favorites, or Quit): ";
std::cin >> command;
if(command == "Create"){
create(&user);
} else if(command == "Update"){
update(&user);
} else if(command == "View"){
view(&user);
} else if(command == "Favorites"){
//favorites(&user);
} else if(command == "Quit"){
break;
} else std::cout << "Please input a valid command.\n";
}
}
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当您取消引用时,u->favs您会获得一个std::string&,并且基于范围,然后开始一次迭代该string一个字符.这就是为什么当您auto从第一个字符串中看到单个字符时,以及当您尝试将该字符绑定到a时,string&您会收到编译错误.
您无法获得基于指针类型的范围,因为它无法知道指针指向的对象数量.如果对代码进行以下更改以声明favs为静态大小的数组,则代码将起作用.
将您的班级定义更改为
using namespace std::string_literals; // allows use of operator""s (user defined literal)
class User
{
public:
User()
: favs{{"Hello"s, "how"s, "are"s, "you"s, "?"s}} // use operator""s
{}
// no need for a destructor definition
std::string lName;
std::string fName;
int age;
std::array<std::string, 5> favs; // statically sized array
};
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如果你需要一个动态大小的数组,那么请std::vector<std::string>改用,你的循环仍然有效.