我正在尝试将这些向量中的一些数据写入文本文件。当我运行代码时,它返回运行时错误。 Category
,Product
,Cart
,Customer
和Address
都是struct
,其成员每个get_member
都会返回。
ofstream write_cats;
write_cats.open("catprd.dat", ios::out, ios::trunc);
vector<Category>::iterator i;
write_cats << cats.size() << endl;
for (i = cats.begin(); i < cats.end(); i++) {
write_cats << i -> get_catid() << '\t';
}
vector<Product>::iterator j;
write_cats << prods.size() << endl;
for (j = prods.begin(); j < prods.end(); j++) {
write_cats << j -> get_prodid() << '\t';
write_cats << j -> get_prodprice() << endl;
}
write_cats.close();
ofstream write_carts;
write_carts.open("carts.dat", ios::out, ios::trunc);
vector<Cart>::iterator k;
write_carts << carts.size() << endl;
for (k = carts.begin(); k < carts.end(); k++) {
write_carts << k -> get_cartid() << '\t';
write_carts << k -> get_day() << endl;
}
vector<Cart_item>::iterator l;
write_carts << cart_items.size() << endl;
for (l = cart_items.begin(); l < cart_items.end(); l++) {
write_carts << l -> get_cartitemid() << '\t';
write_carts << l -> get_qty() << endl;
}
write_carts.close();
ofstream write_custs;
write_custs.open("custs.dat", ios::out, ios::trunc);
vector<Customer>::iterator m;
vector<Address>::iterator n;
write_custs << custs.size() << endl;
for (m = custs.begin(); m < custs.end(); m++) {
write_custs << m -> get_cust_id() << '\t';
write_custs << n -> get_zip_code() << endl;
}
write_custs.close();
返回运行时错误“Vector iterator not dereferencable”
以下是struct Address
的样子:
using namespace std;
#pragma once
#include <string>
struct Address {
public:
int get_st_number() const{return st_number;}
int get_zip_code() const{return zip_code;}
string get_st_name() const{return st_name;}
Address(){}
Address (int num, string name, int zip)
: st_number(num), st_name(name), zip_code(zip) {}
private:
int st_number;
int zip_code;
string st_name;
};
和struct Customer
:
struct Customer {
public:
Address get_address() const{return addr;}
int get_cust_id() const{return cust_id;} customer id
string get_name() const{return cust_name;}
Customer (int id, string n, Address a)
: cust_id(id), cust_name(n), addr(a) {}
string display_addr() const {
std::cout<<setw(15)<<cust_name<<" ";
std::cout<<setw(15)<<cust_id<<" ";
return string();
}
private:
int cust_id;
string cust_name;
Address addr;
};
答案 0 :(得分:3)
您忘了初始化vector<Address>::iterator n;
答案 1 :(得分:1)
您正在声明迭代器n
,但不会将其初始化为可解除引用的值。从您的更新中,您似乎想要打印与客户关联的Address
;所以你可以通过m
引用的客户访问,而不是单独的迭代器:
write_custs << m -> get_cust_id() << '\t';
write_custs << m -> get_address().get_zip_code() << endl;
此外,将每个迭代器的范围放在其循环中可能是个好主意;这比每次在外部范围内声明一个新的更容易出错:
for (vector<Whatever>::const_iterator i = stuff.begin(); i != stuff.end(); ++i) {
// do stuff with "i"
}
// "i" is no longer available - no danger of accidentally using it again.
还有其他几点:
!=
而不是<
来与end()
迭代器进行比较; <
对某些类型的迭代器不起作用; '\n'
而不是endl
; endl
刷新文件缓冲区,强制该文件在那时写入磁盘,这可能非常慢。