c++, using get and >> for ifstream - c++

Question

I have a text file that I am inputting data in from, but I can't seem to get it right.
Here are two lines from the text file as an example (these aren't real people don't worry):
Michael Davidson 153 Summer Avenue Evanston CO 80303
Ingrid Johnson 2075 Woodland Road Aurora IL 60507
Here is the code I have to load the text file and put the data into a struct. I am still new to C++(obviously) and I'm having a hard time using get and >> together. The code I have below, works fine until I get to the "state" and then something goes wrong. Thanks for the help!
//constants
const int FIRST_NAME_LEN = 11;
const int LAST_NAME_LEN = 13;
const int ADDRESS = 25;
const int CITY_NAME_LEN = 16;
const int STATE_LEN = 3;
//define struct data types
struct CustomerType {
char firstName[FIRST_NAME_LEN];
char lastName[LAST_NAME_LEN];
char streetAddress[ADDRESS];
char city[CITY_NAME_LEN];
char state[STATE_LEN];
int zipCode;
};
//prototype function
ifstream& getInfo(CustomerType& CT_Struct, ifstream& infile);
int main() {
//declare struct objects
CustomerType CT_Struct;
ifstream infile("PGM951_customers.txt");
if(!infile) {
cerr << "Could not open the input file." << endl;
exit(1); //terminates the program
}
//call the function
getInfo(CT_Struct, infile);
return 0;
}
ifstream& getInfo(CustomerType& CT_Struct, ifstream& infile) {
while(infile) {
infile.get(CT_Struct.firstName, sizeof(CT_Struct.firstName));
infile.get(CT_Struct.lastName, sizeof(CT_Struct.lastName));
infile.get(CT_Struct.streetAddress, sizeof(CT_Struct.streetAddress));
infile.get(CT_Struct.city, sizeof(CT_Struct.city));
infile.get(CT_Struct.state, sizeof(CT_Struct.state));
infile >> ws;
infile >> CT_Struct.zipCode;
cout << CT_Struct.firstName << " | " << CT_Struct.lastName << " | " << CT_Struct.streetAddress
<< " | " << CT_Struct.city << " | " << CT_Struct.state << " | " << CT_Struct.zipCode << endl;
}
return infile;
}
=== edit ===========
Reading in the state at 8 char was just me messing around and then I forgot to change it back...sorry.

Answer 1

The problem is istream::get() breaks for streetAddress which has spaces in it.
One way is to tokenize the input line first into say, a vector of strings and then depending on the number of tokens convert these to appropriate fields of your CustomerType:
vector<string> tokenize(string& line, char delim=' ') {
vector<string> tokens;
size_t spos = 0, epos = string::npos;
while ((epos = line.find_first_of(delim)) != string::npos) {
tokens.push_back(line.substr(spos, epos - spos));
spos = epos;
}
return tokens;
}
I'd rather a stream extraction operator for CustomerType :
struct CustomerType {
friend istream& operator>>(istream& i, CustomerType& c);
string firstName, lastName, ...;
// ...
};
istream& operator>>(istream& i, CustomerType& c) {
i >> c.firstName >> c.lastName;
string s1, s2, s3;
i >> s1 >> s2 >> s3;
c.streetAddress = s1 + s2 + s3;
i >> c.city >> c.state >> c.zipCode;
return i;
}

Answer 2

You're getting 8 characters for State, which includes all your zipcode, and is larger than your field.
It'd also be tempting to use the skipws operator:
infile >> skipws >> CT_Struct.firstName
>> CT_Struct.lastName
>> ... ;
(Update: that's what I get for doing that from memory. This is more closely approximating correct.)

Answer 3

If I were you I would start again from scratch. I would:
use std::strings instead of character arrays for your data
reads line at a time from the file using std::getline
parse the line up using a stringstream
avoid mixing formatted and unformatted input

Answer 4

My approach to this would be the following:
1) Read each line into a null terminated buffer.
2) Use a split() function that you're gonna have to write. This function should take a string as its input and return a list. It should also take a separator. The separator in this case is ' '.
3) iterate over the list carefully (are there never middle names?) What about 1 word, or 3 word street names? Since many of these columns are really variable in number of words, and you have no seperator other than whitspace, this may prove a fairly tough task. If you NEVER have middle names, you could assume the first two columns are first and last name. You know for sure what the last two are. Everything between them could be assigned to a single address field.