C / C++ / MFC

Re: Getting 2203 error while doing silent installation of MSI

30-Mar-20 3:36

See some of these suggestions: MSI am getting 2203 error - Google Search[^]

Dave Kreskowiak30-Mar-20 3:51

Dave Kreskowiak

30-Mar-20 3:51

2203 in an administrative install usually means "access denied".

Where are you installing it to and is the account running the MSIEXEC command an administrator? Is the .MSI copied to the local machine before you launch it?

Also, "ALLUSERS=0" isn't really a thing. The behavior of ALLUSERS isn't defined for a value of 0. The behavior is defined for values 1 and 2 and for an empty string ("").

A value of 1 is for a "per-machine" install or "all users", which is typical if you're installing to Program Files.

A value of 2 is for letting MSIEXEC determine which type of installation it should do, checking the value of the MSIINSTALLPERUSER property.

See ALLUSERS property - Win32 apps | Microsoft Docs[^] for more.

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Dave Kreskowiak

Re: Getting 2203 error while doing silent installation of MSI

Vijay53331-Mar-20 0:33

Vijay533

31-Mar-20 0:33

Hi Dave,

Thanks for the posting.

1.) I am trying to install in C:\Program Files(x86)\ folder.
2.) I logged in as non admin user, am trying to install the MSI with domain Administrator credentials am using CreateProcessWithLogonW to do this.
3.) Yes, i copied the installer into temp location.

Re: Getting 2203 error while doing silent installation of MSI

Dave Kreskowiak31-Mar-20 4:55

Dave Kreskowiak

31-Mar-20 4:55

OK.

In that case, if the command line your code generated works if an admin on the machine runs it, the problem is going to be in your code that logs in as an admin, but apparently doesn't run anything as an admin. I couldn't tell you what the problem is there.

Asking questions is a skill
CodeProject Forum Guidelines
Google: C# How to debug code
Seriously, go read these articles.

Dave Kreskowiak

Finding a directory

Anthony Appleyard29-Mar-20 18:32

Anthony Appleyard

29-Mar-20 18:32

Thanks for your help to people who ask queries.

I know how to ask for a file to read from, or a file to write to :-

BOOL GetFileName(HWND wn,char*file,char*titl,char*filter,short write/*=1*/) {short j;
strcpy(file,title);
ofn.lStructSize=sizeof(OPENFILENAME); ofn.hwndOwner=wn;
ofn.lpstrFilter=filter; ofn.lpstrCustomFilter=0;
ofn.lpstrFile=file; ofn.nMaxFile=1024;
ofn.lpstrFileTitle=title; ofn.nMaxFileTitle=1024;
ofn.lpstrInitialDir=0; ofn.lpstrTitle=titl;
ofn.Flags=write?OFN_OVERWRITEPROMPT:OFN_FILEMUSTEXIST;
j=(write?GetSaveFileName(&ofn):GetOpenFileName(&ofn))?1:0;
return j;}

('ofn' and 'title' are declared globally.)

but when I need to ask for a directory / folder, I must do something like this:

K=GetFileName(wn, slidefilename, "name of a file in the directory to read slides from?", slidefiletype, 0);

Is there something like GetOpenFileName, but returning the name of a directory rather than the name of a file?

modified 30-Mar-20 0:49am.

Re: Finding a directory

leon de boer29-Mar-20 20:26

leon de boer

29-Mar-20 20:26

SHBrowseForFolder
SHBrowseForFolderA function (shlobj_core.h) - Win32 apps | Microsoft Docs[^]

CodeProject has a sample app
How to Browse for a Folder[^]

In vino veritas

Passing an array as argument to a function

Calin Negru28-Mar-20 22:15

28-Mar-20 22:15

Hi What is the syntax when you want to pass an array as argument? I`m looking for syntax for both function call and function definition.

Richard MacCutchan28-Mar-20 22:34

28-Mar-20 22:34

Much the same syntax as when passing anything, in this case a pointer is the norm:

C++

int myFunction(char* anArrayOfChars) // a pointer to the array
{
    // function code ...
}

...

// calling code
char anArray[32];
// fill the array
int value = myFunction(anArray) // name of the array is a pointer to its content

Calin Negru28-Mar-20 22:59

28-Mar-20 22:59

Thanks
will modifying the array content within the function make the edit permanent i.e. the modification will remain after the function ceased to exist. i.e. "myFunction(&anArray)" won`t this edit outside the function. I need the edit to work backwards.

modified 29-Mar-20 5:32am.

Richard MacCutchan28-Mar-20 23:23

28-Mar-20 23:23

Yes, if the function changes anything in the array then that data will remain when the function returns to the caller. Note: you do not need the addressof operator (&) on an array name. Just code it as I showed in my example.
Try this simple test:

C++

void myFunc(char* somedata)
{
    strcpy(somedata, "Bad stuff");
    return;
}

//...
char myArray[] = "Good stuff";
myFunc(myArray);
printf("Returned value %s\n", myArray);

Calin Negru28-Mar-20 23:44

28-Mar-20 23:44

I edited my previous post a bit

also

void myFunc(char* somedata)
{
    strcpy(somedata, "Good stuff");
    return;
}

//...
char myArray[] = "Bad stuff";
myFunc(myArray);
printf("Returned value %s\n", myArray);

Quote:
void myFunc(char* somedata)
{
strcpy(somedata, "Good stuff");
return;
}

//...
char myArray[] = "Bad stuff";
myFunc(myArray);
printf("Returned value %s\n", myArray);

will this function? the array has 9 elements(bad stuff 9 chars.) "good stuff" is 10.

modified 29-Mar-20 6:10am.

Richard MacCutchan28-Mar-20 23:55

28-Mar-20 23:55

Well, of course that breaks it so you may find that your code crashes. You must not overwrite arrays that are passed to you with more data than it can hold. But this conversation is missing some vital information. Maybe you can explain exactly what problem you are trying to solve.

Calin Negru29-Mar-20 0:48

29-Mar-20 0:48

@Richard Meh I`m really not doing anything worth much attention. My code is stashed with hacks so I`m just trying to fix on it.

"DreamLand Page" on facebook

Richard MacCutchan29-Mar-20 0:50

29-Mar-20 0:50

fearless_ wrote:
My code is stashed with hacks

So it all needs a lot of attention.

Calin Negru29-Mar-20 1:14

29-Mar-20 1:14

Thanks for pointing how things are meant to be done Richard

"DreamLand Page" on facebook

k505429-Mar-20 4:03

k5054

29-Mar-20 4:03

Further to what Richard has said, you could also do some defensive programming by passing in the length of the array. eg:

C++

void myFunc(int* somedata, size_t data_len)
{
    for(size_t i = 0; i < data_len; ++i)
        somedata[i] *= 2;
    return;
}

//...
int myArray[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10);
const size_t arrLen = sizeof(myArray)/sizeof(myArray[0]);
myFunc(myArray), arrLen);
// ... etc

This will prevent data overflows, and/or segfaults, when accessing myArray inside the function.
Things to note:
* we can get the compiler to tell us how many elements in the array using the sizeof(myArray)/sizeof(myArray[0]) construct. This is calculated at compile time and in release mode, in most cases it will be optimized out, so it does not add to your memory usage, if that's a concern.
* Using the constant arrLen means that if we change the number of elements in myArray, we don't need to go through the code and find all uses of myArray and make sure we're passing in the right number
* If we had used arrLen = sizeof(myArray)/sizeof(int), we need to remember to change the definition of arrLen if we change the type of myArray. Using sizeof(myArray[0] mean that if we change the type of myArray from int[] to double[], for example, we don't need to remember to change the definition of arrLen as well

Keep Calm and Carry On

Richard MacCutchan29-Mar-20 4:18

29-Mar-20 4:18

There is also the _countof() macro, which gives the number of elements in arrays of any simple or composite type.

leon de boer29-Mar-20 4:23

leon de boer

29-Mar-20 4:23

Best macro ever added into C standard ... wish more people would use it.

In vino veritas

Richard MacCutchan29-Mar-20 4:28

29-Mar-20 4:28

They probably would if it was easier to find; after all it is in the documentation. Wink | ;)

k505429-Mar-20 4:34

k5054

29-Mar-20 4:34

I didn't know about _countof(). Is it an MS only extension? Trying to compile with gcc under linux produces an implicit declaration warning in C and a not declared in this scope in C++

Keep Calm and Carry On

Richard MacCutchan29-Mar-20 5:48

29-Mar-20 5:48

The documentation on MSDN (_countof Macro | Microsoft Docs[^]) does not identify it as MS only.