Container of buffers

[Alex Vinokur]

I need a container of char-buffers, for instance, vector of
char-buffers.

vector<string> v; // It is not what I need to

char a[10];
memset (&a[0], 0, sizeof (a));

v.push_back(a);

cout << v.back(); // This displays empty string, not 10 bytes. I need
to get 10 bytes in this example.

Is there any way to create a container of char-buffers?
Alex Vinokur
email: alex DOT vinokur AT gmail DOT com
http://mathforum.org/library/view/10978.html
http://sourceforge.net/users/alexvn

[Jakob Bieling]

Alex Vinokur <al****@users.sourceforge.net> wrote:

I need a container of char-buffers, for instance, vector of
char-buffers.

vector<string> v; // It is not what I need to
I would say this would work, if used right. But there are other
solutions.
char a[10];
memset (&a[0], 0, sizeof (a));

v.push_back(a);
Because v is a vector of strings, a temporary std::string object is
constructed based on 'a'. That means 'a' is treated like a C-style
string. Because 'a' contains all '\0' values, the char-array contains an
empty C-string, which is why the std::string object is also empty.
cout << v.back(); // This displays empty string, not 10 bytes. I need
to get 10 bytes in this example. Is there any way to create a container of char-buffers?

std::vector <std::string> v;
v.resize (20, std::string (10, 0));

Now 'v' contains 20 strings. Each of those contains 10 0-chars. Note
that you can dynamically change the size of each string. Instead of
using a std::string, you could also use a std::vector.

Another approach is to create a struct:

struct char_buffer
{
char data [10];
};

std::vector <char_buffer> v;
v.resize (20);
Now 'v' contains 20 'char_buffer' objects. Each of those contains 10
chars. Note that now you cannot dynamically change the size of each
char_buffer anymore.

hth
--
jb

(reply address in rot13, unscramble first)

[flagos]

Alex. Try something like the following:

typedef std::vector<std::string> my_container;

my_container c;

std::string s1(10,' ');

c.push_back( s1 );
Hope this help.