From: "Matt Dooner" <mdooner AT mindspring DOT com>
Newsgroups: comp.os.msdos.djgpp
Subject: Re: gpp can't find headers
Date: Sun, 9 Apr 2000 05:53:11 -0400
Organization: MindSpring Enterprises
Lines: 54
Message-ID: <8cpjtm$okd$1@slb0.atl.mindspring.net>
References: <8codd9$equ$1 AT slb6 DOT atl DOT mindspring DOT net> <i4f0fssllor07t7n7u7atlfefd53ftq30a AT 4ax DOT com>
NNTP-Posting-Host: cf.45.30.6e
X-Server-Date: 9 Apr 2000 09:52:54 GMT
X-Newsreader: Microsoft Outlook Express 4.72.3155.0
X-MimeOLE: Produced By Microsoft MimeOLE V4.72.3155.0
To: djgpp AT delorie DOT com
DJ-Gateway: from newsgroup comp.os.msdos.djgpp
Reply-To: djgpp AT delorie DOT com

The program I posted was just a test program.   The bad code is what happens
when I put together something quickly with a combination of jet lag and lack
of sleep :-)

Jason Green wrote in message ...
>"Matt Dooner" <mdooner AT mindspring DOT com> wrote:
>
>> I've followed the instructions in the FAQ, but gpp still can't find my
>> included headers in c++ programs.   Here is the error message from a
basic
>> program that uses a string and vector.   Below it is the code of the
>> program;
>
>> #include <iostream.h>
>> #include <string.h>
>> #include <vector.h>
>
>These are old style headers.  You need to use the new style if you
>want to use the string class:
>
>#include <iostream>
>#include <string>
>#include <vector>
>using namespace std;
>
>>
>> int main()
>> {
>>   vector test<int>;
>
>  vector<int> test;
>
>>   test.resize(1);
>>   string test = "test";
>
>test is already defined as type vector<int> so you can not define
>another variable of type string with the same name.  It is not clear
>what you want to do here, if you wanted a vector of strings then test
>should have been defined as type vector<string>.  If you want a second
>variable of type string then it must have a unique name:
>
>  string str = "test";
>
>>   cout << test << endl;
>
>You can not just output a vector to cout.  You must iterate through
>each value contained in the vector and output them individually.  You
>can still output the string though:
>
>  cout << str << endl;
>
>> }