Printing to Default Printer

[mmscg]

In the simple program below, what do I have to do to send the output
to the default printer rather than a message box?

#include <windows.h>


int WINAPI WinMain(HINSTANCE hinst, HINSTANCE hinstPrev, LPSTR lpszCmdLine, int nCmdShow)
{
    int i;

    for (int i=1; i < 6; i++)
    {
	MessageBox(NULL, i, NULL, MB_ICONWARNING);
    }

	return 0;
}

Printed page output like so:

1
2
3
4
5

Seems trivial I know, and yes I've searched, but I cannot find what I am looking for.

[Marc G]

Maybe the following articles can get you started: http://www.codeguru.com/cpp/w-p/printing/

[mmscg]

Thanks for link.

Those are a bit more complicated than I was hoping for, but if the only way,
I will try to understand.

In the mean time, I was hoping for a one line printer call

I found these 2 simple snippets, but when I insert either into my program it won't compile.

Any ideas what I am doing wrong?

fprintf(stdprn,"Test test test\f");

FILE *stdprn;
stdprn = fopen("PRN","wb");
fprintf(stdprn,"From Windows: Test test test\f");
fclose(stdprn);

[MrViggy]

Define "won't compile". What are the errors?

Viggy

[mmscg]

OK, I finally got this to work (not to printer, but a file).

I tested on two different computers.

0 errors, 0 warnings on both computers.

#include <windows.h>
#include <stdio.h>

int WINAPI WinMain(HINSTANCE hInstance , HINSTANCE hPrevInstance ,
			       PSTR lpCmdLine , int nCmdShow ) 
{
    
    FILE *fp;
    char *str; 
 
    fp = fopen("sample.txt", "w");

    str = "line 1\n";
    fwrite(str, 1, strlen(str), fp);
 
    str = "line TWO\n";
    fwrite(str, 1, strlen(str), fp);

    MessageBox(NULL , TEXT("File Written") , lpCmdLine , MB_OK);
    
    return 0;
}

Contents of sample.txt for both computers.

line 1
line TWO

Contents of debug window on one computer

'Print.exe': Loaded 'C:\... ...\WriteFile00\Debug\Print.exe', Symbols loaded.
'Print.exe': Loaded 'C:\WINDOWS\system32\ntdll.dll', No symbols loaded.
'Print.exe': Loaded 'C:\WINDOWS\system32\kernel32.dll', No symbols loaded.
'Print.exe': Loaded 'C:\WINDOWS\system32\user32.dll', No symbols loaded.
'Print.exe': Loaded 'C:\WINDOWS\system32\gdi32.dll', No symbols loaded.
'Print.exe': Loaded 'C:\WINDOWS\system32\uxtheme.dll', No symbols loaded.
'Print.exe': Loaded 'C:\WINDOWS\system32\msvcrt.dll', No symbols loaded.
'Print.exe': Loaded 'C:\WINDOWS\system32\advapi32.dll', No symbols loaded.
'Print.exe': Loaded 'C:\WINDOWS\system32\rpcrt4.dll', No symbols loaded.
The program '[1316] Print.exe: Native' has exited with code 0 (0x0).

Contents of debug window on another computer

'Print.exe': Loaded 'C:\... ...\WriteFile00\Debug\Print.exe', Symbols loaded.
'Print.exe': Loaded 'C:\WINNT\system32\NTDLL.DLL', Cannot find or open a required DBG file.
'Print.exe': Loaded 'C:\WINNT\system32\USER32.DLL', Cannot find or open a required DBG file.
'Print.exe': Loaded 'C:\WINNT\system32\KERNEL32.DLL', Cannot find or open a required DBG file.
'Print.exe': Loaded 'C:\WINNT\system32\GDI32.DLL', Cannot find or open a required DBG file.
The program '[624] Print.exe: Native' has exited with code 0 (0x0).

Why the different output?

What does Cannot find or open a required DBG file mean?

[S_M_A]

What does Cannot find or open a required DBG file mean?

It just means that those dll's don't have any debug information, i.e. nothing to be worried about.

[mmscg]

OK, Thanks!

I can continue on this path then.

[mmscg]

Moving forward, I have this:

#include <windows.h>
#include <stdio.h>

int WINAPI WinMain(HINSTANCE hInstance , HINSTANCE hPrevInstance ,
			       PSTR lpCmdLine , int nCmdShow )
{
    
    FILE *fp;
    char *str;
    char buffer [33];

    fp = fopen("sample.txt", "w");

    str = "output:\n";
	fwrite(str, 1, strlen(str), fp);

    for (int i=1; i < 6; i++)
    {
		
		switch ( i )
		{
			case 1:
				str = "(1st line)\n";
				break;
			case 2:
				str = "(2nd line)\n";
				break;
			case 3:
				str = "(3rd line)\n";
				break;
			case 4:
				str = "(4th line)\n";
				break;
			case 5:
				str = "(5th line)\n";
				break;
		}		

		fwrite("i = ",1,4,fp);
		fwrite(itoa(i, buffer, 10), 1, 1, fp);
		fwrite("   ", 1, 3, fp);
		fwrite(str, 1, strlen(str), fp);

    }
   
    MessageBox(NULL , TEXT("File Written") , lpCmdLine , MB_OK);
    
    return 0;
}

Contents of sample.txt:

output:
i = 1   (1st line)
i = 2   (2nd line)
i = 3   (3rd line)
i = 4   (4th line)
i = 5   (5th line)

Is it possible to replace the four fwrites with one, ang get the same output?

Something like?

fwrite("i = " & i & " " & str, 1, 18, fp);

[Paul McKenzie]

const char *sArray[] = { 
    "(1st line)\n",
    "(2nd line)\n",
    "(3rd line)\n",
    "(4th line)\n",
    "(5th line)\n"
};
//..
for (int i = 0; i < sizeof(sArray) / sizeof(sArray[0]); ++i )		
{
   // no case statement needed now
   str = sArray[i];
   //... 
}

Regards,

Paul McKenzie

[mmscg]

Thanks Paul for the reply.

I was not trying to get rid of the switch block, but rather have one fwrite statement to accomodate different data types; strings, integers, etc.

I found that the fprintf statement would give me what I was after.

#include <windows.h>
#include <stdio.h>

int WINAPI WinMain(HINSTANCE hInstance , HINSTANCE hPrevInstance ,
			       PSTR lpCmdLine , int nCmdShow )
{
    
    FILE *fp;
    char *str;
    byte b;

    b=255;
    fp = fopen("sample.txt", "w");

    str = "output:\n";
    fwrite(str, 1, strlen(str), fp);

    for (int i=1; i < 6; i++)
    {
		
		switch ( i )
		{
			case 1:
				str = "(1st line)";
				b=34;
				break;
			case 2:
				str = "(2nd line)";
				b=127;
				break;
			case 3:
				str = "(3rd line)";
				b=122;
				break;
			case 4:
				str = "(4th line)";
				b=13;
				break;
			case 5:
				str = "(5th line)";
				b=94;
				break;
		}		

		fprintf(fp,"i = %d %s %x\n", i, str, b);

    }
   
    MessageBox(NULL , TEXT("File Written") , lpCmdLine , MB_OK);
    
    return 0;
}

sample.txt file

output:
i = 1 (1st line) 22
i = 2 (2nd line) 7f
i = 3 (3rd line) 7a
i = 4 (4th line) d
i = 5 (5th line) 5e

Seems OK.

Now to try to put this in my real program, and see if it gives me the output I am after.

[mmscg]

I think I have the "printing" working correctly,
but I find some strange results when looking over the printed output.

Is it a problem with my fprintf usage?

In the output below generated by the following code,
why the 8 characters «««««««« following untitled?

Code:

//-------------------------------------------------------
LPVOID Alloc(DWORD dwSize)
{
    return HeapAlloc(g_hheap, HEAP_ZERO_MEMORY, dwSize);
}
//-------------------------------------------------------


//-------------------------------------------------------
struct EVENT {
    int     nData;
    LPBYTE  lpData;
};

typedef struct EVENT EVENT;
typedef struct EVENT *LPEVENT;

LPEVENT lpEvent;
DWORD dw;

dw = 8;
lpEvent->nData  = dw;
lpEvent->lpData = (LPBYTE)Alloc(lpEvent->nData);
mmioRead(hmmio, (HPSTR)lpEvent->lpData, lpEvent->nData);

fprintf(fp,"lpEvent->nData   = %d\n", lpEvent->nData);
fprintf(fp,"lpEvent->lpData  = <%s>\n", lpEvent->lpData);
//-------------------------------------------------------

Output:

lpEvent->nData   = 8
lpEvent->lpData  = <untitled««««««««>

[Marc G]

Because lpEvent->lpData is not NULL terminated.
If you want to read 8 characters from a file into a buffer and later print it, it's best to allocate 9 bytes, read 8 bytes from file and set the 9th byte to \0. That way, fprintf will stop correctly at the \0 and will not print garbage.

[mmscg]

I think I understand what you are saying, and I will try as you suggest.

What I do not understand is the length of my returned string; 8 characters in untitled and 8 «

Is lpEvent->lpData a pointer to a buffer?

If yes, does this buffer have a fixed or limiting size?

[Marc G]

Yes, lpEvent->lpData is some kind of buffer.
When you pass it to fprintf, fprintf will just output characters until it finds a \0 character.

[mmscg]

With regards to the code/output in Post #11, I would like output as so:

lpEvent->nData   = 8
lpEvent->lpData  = <untitled««««««««>
char 1 = u = hex = 75
char 2 = n = hex = 6E
char 3 = t = hex = 74
char 4 = i = hex = 69
char 5 = t = hex = 74
char 6 = l = hex = 6C
char 7 = e = hex = 65
char 8 = d = hex = 64

This now is not a question of how to use the fprintf function, but rather
how to read each character in turn from the lpEvent->lpData string and convert to hex.