Re: [suse-programming-e] reading a file and display its content with cout
do you mean like this:
"
char mybuffer[];
fstream sumber("smb.conf");
if (!sumber){
cout << "Failed in opening the 'smb.conf'
file\n";
exit(1);
}
sumber >> mybuffer;
cout << mybuffer << endl;
"?
Please tell me more details. Could you give me a very
simple sample?
Thank you very much for your generousity.
--- Alan Lenton
Dear my friend.
I want to make a linux server administrating tool ( I am also a beginner in GNU C++).
if I open the file (smb.conf) with ifstream and
Prabu Subroto wrote: try to
display its content with cout, but it does not display its content only somewhat looksline only an address of the variable in the memory. like this: " psubroto@vh64:~/arsip/proyek/g++/BLiSS> ./liss Choose the server you would like to administrate : 1. Samba Server 1 Samba Server Administrating Menu --------------------------------
Workgroup name :bitp nama workgroup-nya : bitp 0xbfffef64 psubroto@vh64:~/arsip/proyek/g++/BLiSS> "
what is my mistake? [snip...]
void bsamba::menu(){ cout << "Samba Server Administrating Menu\n"; cout << "--------------------------------\n" << endl; cout << "Workgroup name :"; cin >> bwg;
cout << "nama workgroup-nya : " << bwg << endl; fstream sumber("smb.conf"); if (!sumber){ cout << "Failed in opening the 'smb.conf' file\n"; exit(1); } cout << sumber << endl; }
[snip...]
What you have asked it to print out is not the file, but the address of the stream. You need to use the stream (sumber) to read the file into a buffer and then write the buffer out to cout :)
alan
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do you mean like this: " char mybuffer[]; fstream sumber("smb.conf"); if (!sumber){ cout << "Failed in opening the 'smb.conf' file\n"; exit(1); } sumber >> mybuffer; cout << mybuffer << endl; "?
Please tell me more details. Could you give me a very simple sample? yes, but you are not allocating space for mybuffer. You must declare a size. In the case below, I create a file, test.dat, but you could use smb.conf Nore that the >> operator reads a word. If you want to read a line, use the getline() member function. Here is an example, below. #include <iostream> #include <fstream> using std::cin; using std::cout; using std::cerr; using std::fstream; using std::endl; using std::ios;
On Monday 31 January 2005 10:00, Prabu Subroto wrote:
main ()
{
fstream file1;
char mybuffer[512];
file1.open( "/etc/samba/smb.conf", ios::in);
if (!file1)
{
cerr << "Unable to open test.dat." << endl;
exit(1);
}
while( !file1.eof() ) {
file1.getline(mybuffer, sizeof(mybuffer));
cout << mybuffer << endl;
}
file1.close();
}
--
Jerry Feldman
participants (2)
-
Jerry Feldman -
Prabu Subroto