Re: CPrintDialog default printer. - C / C++ / MFC Discussion Boards

Re: Can GDI and RichEdit share the WindowClient (and live together)

ForNow28-Nov-21 8:33

ForNow

28-Nov-21 8:33

thanks going try one more thing before bypassing richedit's coloring capabilities

Please give me explanation to this code.Explain me the second line of code. What will be the output?

poojitha0822-Nov-21 19:04

poojitha08

22-Nov-21 19:04

# include <stdio.h>
# define scanf "%s me "
main()
{
printf(scanf, scanf);
getchar();
return 0;
}

Re: Please give me explanation to this code.Explain me the second line of code. What will be the output?

Victor Nijegorodov22-Nov-21 20:31

Victor Nijegorodov

22-Nov-21 20:31

Didn't you try to compile/test it? Confused | :confused:

[REPOST] Please give me explanation to this code.Explain me the second line of code. What will be the output?

Richard Deeming22-Nov-21 21:31

Richard Deeming

22-Nov-21 21:31

You already have the answer on your repost of this:
Please give me explanation of this code.[^]

"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer

CPrintDialog default printer.

Maximilien18-Nov-21 3:40

Maximilien

18-Nov-21 3:40

I don't have a physical printer, so I cannot test (Am I wrong here?)

I'm trying to print something (a bitmap) using MFC.

I set my default printer to "Microsoft Print To PDF" in Windows 10 settings.

I have a skeleton code like this :

The GetDefaults() fails.

Am I missing something ?

Obviously, other print commands in other software work (VS, Word, Chrome... )

{
    CPrintDialog dlg(FALSE);

    if (!dlg.GetDefaults())
    {
        AfxMessageBox(_T("You have no default printer!"));
    }
    else
    {
        // attach to the DC we were given
        CDC dc;
        dc.Attach(dlg.m_pd.hDC);
        //....
    }

}

CI/CD = Continuous Impediment/Continuous Despair

Re: CPrintDialog default printer.

Victor Nijegorodov18-Nov-21 7:17

Victor Nijegorodov

18-Nov-21 7:17

You don't need the CPrintDialog if you already know the printer name.
Just use something like

C++

HANDLE printerHandle = nullptr;
TCHAR name[] = _T("Microsoft Print To PDF");
if (!::OpenPrinter(name, &printerHandle, nullptr))
{
   // handle the error
   ...
}

Re: CPrintDialog default printer.

Maximilien19-Nov-21 5:00

Maximilien

19-Nov-21 5:00

the problem is not the device name, but GetDefaults not returning the default printer.

Obviously, it should work for any device.

CI/CD = Continuous Impediment/Continuous Despair

Re: CPrintDialog default printer.

Randor 19-Nov-21 12:34

Randor

19-Nov-21 12:34

Max,

The default constructor is in the documentation[^]:

C++

CPrintDialog(
    BOOL bPrintSetupOnly,
    DWORD dwFlags = PD_ALLPAGES | PD_USEDEVMODECOPIES | PD_NOPAGENUMS | PD_HIDEPRINTTOFILE | PD_NOSELECTION,
    CWnd* pParentWnd = NULL);

The default constructor does not contain PD_RETURNDEFAULT. You will need to use the PD_RETURNDEFAULT flag[^] to get the CPrintDialog class to simply return the default printer. In fact by using this flag the DoModal(); function won't even open a window.

C++

CPrintDialog dlg(FALSE,PD_RETURNDEFAULT,NULL);
dlg.DoModal();
LPDEVMODE lpdev = dlg.GetDevMode();
MessageBox(lpdev->dmDeviceName,0,0);

Windows 10 makes decisions for the novice users. So Windows 10 will sometimes decide which printer to use and override the user choice. To disable this programmatically you could set the HKCU\SOFTWARE\Microsoft\Windows NT\CurrentVersion\Windows\LegacyDefaultPrinterMode to TRUE.

Best Wishes,
-David Delaune

Re: CPrintDialog default printer.

Maximilien22-Nov-21 1:48

Maximilien

22-Nov-21 1:48

lol, thanks.

I should really pickup this technical skill of RTFM.

CI/CD = Continuous Impediment/Continuous Despair

Re: CPrintDialog default printer.

Randor 22-Nov-21 2:45

Randor

22-Nov-21 2:45

You are welcome.

++ operator in the case of pointers

Calin Negru15-Nov-21 5:32

Calin Negru

15-Nov-21 5:32

I have some basic knowledge about pointers and I`m trying to get an exhaustive understanding, like an all sides perspective. My question is what does the operator ++ do to a pointer? I`ve seen it being used in linked lists related code.

Re: ++ operator in the case of pointers

Richard MacCutchan15-Nov-21 6:20

Richard MacCutchan

15-Nov-21 6:20

It is the increment operator, and as with any variable it adds 1 to the pointer. This will move the pointer forward by one element of whatever it is pointing to: int, long, struct, object etc.

Re: ++ operator in the case of pointers

Calin Negru15-Nov-21 7:10

Calin Negru

15-Nov-21 7:10

could I increment not the value of the variable the pointer is pointing to but the pointer itself?

int Something0 = 0;
int Something1 = 1;
int ** Pointers = new int*[2];
Pointers[0] = &Something0;
*Pointers[0]++; //increase Something0 by 1
Pointers[1] = &Something1;
*Pointers[1]++; //increase Something1 by 1

//Could I do this 
**Pointers =10; //set Something0 to 10
  Pointers++;
**Pointers =30;  //set Something1 to 30

Re: ++ operator in the case of pointers

Richard Andrew x6415-Nov-21 7:26

Richard Andrew x64

15-Nov-21 7:26

That's what the increment operator does to a pointer - it increments the pointer itself.

If you want to increment the value that it's pointing to, you must dereference the pointer first, like so: (*pointer)++;

The difficult we do right away...
...the impossible takes slightly longer.

Re: ++ operator in the case of pointers

Calin Negru15-Nov-21 7:38

Calin Negru

15-Nov-21 7:38

Thanks I think I understand.

Re: ++ operator in the case of pointers

k505415-Nov-21 11:37

k5054

15-Nov-21 11:37

You should also know about the difference between a pre-increment and post-increment operator. In an expression, a pre-increment operator computes the increment before accessing the item, and a post-increment computes the increment afterwards. For example

C++

int i = 1;
int j = ++i;   // i is incremented before assignment: i = 2, j = 2;
int k = i++;   // i is incremented after assignment: i = 3, k = 2;
int x = ++i++; // UNDEFINED BEHAVIOR!!! Do not do this!!!

In the case ++i++ The standard does not say what will happen. The only thing you know is that i will have the value 5 after the expression is complete. The variable x could be assigned at any point during the expression, or since this is undefined behavior, it might even get the value -42, which makes no sense in respect to the code, but is acceptable according to C/C++ standard. Even if you do some testing and work out that for type int the expression ++i++ is the equivalent of ++(i++), it might be different for type long, or for a pointer type, and it might change depending on complexity of the overall expression, and/or the optimization level. The takeaway for this is that you don't know what will happen if you try to pre and post increment an expressions term at the same time. Moreover, you should not try to increment the same variable within the same expression:

C++

int i = 2;
printf("%d %d\n", i++, ++);  // undefined order of evaluation

In the printf statement above, its undefined which order the arguments to printf get evaluated, and different compilers do it differently. I'm reasonably sure that clang and gcc do this differently.

Keep Calm and Carry On

Re: ++ operator in the case of pointers

Calin Negru16-Nov-21 2:55

Calin Negru

16-Nov-21 2:55

thanks k5054. I`m not sure I fully understand the difference. I guess I need look up the issue in several different sources. (I often find consulting different perspectives on the same issue helpful).

modified 16-Nov-21 17:41pm.

Re: ++ operator in the case of pointers

Richard MacCutchan17-Nov-21 3:02

Richard MacCutchan

17-Nov-21 3:02

You should also read Why Does x = ++x + x++ Give Me the Wrong Answer?[^].

Re: ++ operator in the case of pointers

Calin Negru17-Nov-21 6:10

Calin Negru

17-Nov-21 6:10

Thanks

Re: ++ operator in the case of pointers

jsc4217-Nov-21 4:58

jsc42

17-Nov-21 4:58

I am guessing that

k5054 wrote:

C++

int i = 2;
printf("%d %d\n", i++, ++);  // undefined order of evaluation

should have read as

C++

int i = 2;
printf("%d %d\n", i++, ++i);  // undefined order of evaluation

Re: ++ operator in the case of pointers

k505417-Nov-21 5:20

k5054

17-Nov-21 5:20

Yes, of course. My bad.

Keep Calm and Carry On

c++ code interpreter/debugging tool

Calin Negru10-Nov-21 1:40

Calin Negru

10-Nov-21 1:40

A while ago I`m started making a c++ code debugger tool in Forms (C#). The tool is designed to spot faults in a c++ code file and works such that in the process no compilation is taking place. It`s basically a simulation of execution without a binary file being involved. Are there similar projects around?

Re: c++ code interpreter/debugging tool

jeron110-Nov-21 6:09

jeron1

10-Nov-21 6:09

CalinNegru(fearless_) wrote:
It`s basically a simulation of execution without a binary file being involved

Sorry, I don't have an answer for you, but wow, that's sounds incredibly ambitious. OMG | :OMG:

"the debugger doesn't tell me anything because this code compiles just fine" - random QA comment
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"I don't drink any more... then again, I don't drink any less." - Mike Mullikins uncle

Re: c++ code interpreter/debugging tool

Calin Negru10-Nov-21 7:36

Calin Negru

10-Nov-21 7:36

It`s similar to what Visual Studio Intellisense does just driven one step further

Re: c++ code interpreter/debugging tool

jeron110-Nov-21 7:47

jeron1

10-Nov-21 7:47

Simulating execution versus providing context sensitive help seems like a BIG step, at least to me. I'm not dismissing Intellisense, which I imagine is a very large and complex task itself.

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