Why is this returning an address in the console?

Question

I'm trying to wrap my head around ostringstreams and istringstreams. So, as I always do, I made a log-in program out of it. but every time I try to cout the contents of the username and password variables, it returns the address!

Purpose for program: to create a simulated log-in screen using input and output stringstreams

code:

#include<iostream>
#include<string>
#include<conio.h>
#include<stdio.h>
#include<sstream>

using namespace std;

int main(int argv, char *argc[]){

char ch;
ostringstream username,
    password;
ostringstream *uptr, 
    *pptr;

uptr = &username;
pptr = &password;

cout << "Welcome" << endl << endl;

cout << "Enter a username: ";
do{

    ch = _getch();
    *uptr << ch;
    cout << ch;

}while(ch != '\r');


cout << endl << "Enter a Password: ";
do{
    ch = _getch();
    *pptr << ch;
    cout << "*";

}while(ch != '\r');

//if(username == "graywolfmedia25@gmail.com" && password == "deadbeefcoffee10031995"){
    cout << endl << "username: " << *username << endl << "password: " << *password << endl;
//} else {
    //cout << endl << "ACCESS DENIED" << endl;
//}



return 0;
}

I tried using the *uptr and *pptr last, but before that I tried just writing and reading straight from the variables.

@BryanChen you are a beautiful person XD thank you for the help. and sorry for the unclear code.. i'm working on it :P can you explain the .str() class? or link me to a good resource that will? — Mitchell Gray Edwards, Nov 28 '13 at 23:06

score 2 · Accepted Answer · answered Nov 28 '13 at 23:39

2

you should use str to get std::string from ostringstream

so

cout << endl << "username: " << username.str() << endl << "password: " << password.str() << endl;

answered Nov 28 '13 at 23:39

Bryan Chen

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score 1 · Answer 2 · answered Nov 28 '13 at 23:08

The standard streams have output operators for addresses: when you try to print a pointer, it will just print the pointer's address. In addition, the streams have a conversion to pointer which used to indicate if the stream is in a good state: when it is in good state, i.e., stream.fail() == false, it converts to a suitable non-null pointer, normally just this. When it is in failure state it return 0 (the reason it isn't converting to bool is to avoid, e.g., std::cout >> i to be valid: if it would convert to bool this code would be valid).

Assuming you want to print the content of the string stream, you'd just use stream.str() to get the stream's std::string.

Why is this returning an address in the console?

2 Answers2