C / C++ / MFC

9-Feb-06 13:45

I am a great fan of MFC's CString class, especially it's Format member function. Makes things so easy. But I now have a problem that requires me to use a strstream buffer in a loop. How do I go about reinitializing the strstream so that I write to the beginning of the buffer at each loop iteration. This code does not work, the string in the stream does not change at all after the first iteration (stays at "%%1")

ostrstream stream;
for (x = 1; x < 10; ++x)
{
   stream.seekp(0);
   stream << "%%" << x << ends;
...
}

This code however does work

for (x = 1; x < 10; ++x)
   ostrstream stream;
   stream << "%%" << x << ends;
...
}

Recreating the strstream in each loop iteration seems like a silly way to get this to work.

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_anil_9-Feb-06 14:07

9-Feb-06 14:07

use stream.clear(); inside the loop.

PJ Arends9-Feb-06 14:20

9-Feb-06 14:20

Nope, still does not work. I tried clear() alone, clear() followed by seekp(), and seekp() followed by clear(). All the same result as before.

_anil_9-Feb-06 14:35

9-Feb-06 14:35

actually no need of clear also. your code is right.
I want to know how did you get to know that the output is wrong.
please give that part of code.
because every thing is going fine in my code.

PJ Arends9-Feb-06 15:00

9-Feb-06 15:00

cout << stream.str() << endl;

This line is added in the loop for debugging only. and the output is a series of identical lines "%%1". I also looked up the value in the debugger, used the second ostrstream c'tor where you can supply a char * buffer to be used as the stream and watched that buffer, etc. All have the same effect.

I am using VC6 SP6 with the Feb 2001 SDK if that makes any difference. I tried the code with VC7.0 but that was an even worse experience than with VC6, calling stream.str() results in an access violation at address 0x00000000.

Christian Graus is always saying how streams are the best thing ever, but I am having serious doubts about that Sigh | :sigh:

_anil_9-Feb-06 15:34

9-Feb-06 15:34

<code>ostrstream stream;
	for (int x = 1; x < 10; ++x)
	{
		stream.seekp(0);
		stream << "%%" << x << ends;
		CString str = stream.str();
	}
</code>

In the above code I am getting str value %%1, %%2 and so on...
I don't know why you have the problem.

Regards
Anil

PJ Arends9-Feb-06 15:42

9-Feb-06 15:42

#include <iostream>
#include <strstream>

using namespace std;

int main(int argc, char* argv[])
{
    unsigned x;
    ostrstream stream;
    for (x = 1; x < 10; ++x)
    {
        stream.seekp(0);
        stream << "%%" << x << ends;
        cout << stream.str() << endl;
    }

	return 0;
}

output is 9 lines of %%1

I rebooted my computer, created a new project just to test this and yet I still get the same result. Oh well, I guess I will go back to the inefficient way that works for me.

Thanks for your attempt to help.

_anil_9-Feb-06 15:56

9-Feb-06 15:56

Yes I got the solution but don't know the reason.
in VC6 you have to include #include <strstrea.h> instead of #include <strstream> I guess. The folloing code will work.

<code>
#include <iostream>
#include <strstrea.h>  // Hear is the change

using namespace std;

int main(int argc, char* argv[])
{
    unsigned x;
    ostrstream stream;
    for (x = 1; x < 10; ++x)
    {
	stream.seekp(0);
        stream << "%%" << x << ends;
	std::cout << stream.str() << endl;
    }

	return 0;
}
</code>

Regards
Anil

PJ Arends9-Feb-06 16:15

9-Feb-06 16:15

Ok, mystery solved. The older version of strstream as declared in strstrea.h does not call freeze() which is why the code worked for you.

Thanks for your help Big Grin | :-D

Got it!

PJ Arends9-Feb-06 15:53

9-Feb-06 15:53

#include <iostream>
#include <strstream>

using namespace std;

int main(int argc, char* argv[])
{
    unsigned x;
    ostrstream stream;
    for (x = 1; x < 10; ++x)
    {
        stream.seekp(0);
        stream << "%%" << x << ends;
        cout << stream.str() << endl;
        stream.rdbuf()->freeze(false);
    }

	return 0;
}

stream.str() calls freeze(true) to lock the buffer, I have to call freeze(false) to unlock it. Why your code worked without it is a mystery to me.