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David, Let me explain how the path is arrived at. The User selects certain options in a dialog; and upns the path is predetermined, hence the usage of a CString. The path will only be the same is the User makes the same selections. How am I to know what the User wants before he makes an selection? I can't, so I must be prepared to use his selections for the path. Belie ve me it's complicated when you are building a path according to selections made by a User. As you can see I can't do the _T("%s") rutine everytime, in advance. The data being gathered is for different wells and their locations(paths). I no nothing about what you mentioned as "L" or how to do it.
According to Mark CStrings are suppose to be the same as TCHAR's, but my compiler doesn't know that yet!
the code Mark compiled and ran on his machine and he didn't use the _T("%s") or the "L" why won't it on mine? is the question and evidently the resolution.
Of course I further edited it with the _T("%s") and it compiled and ran correctly. But I can't hardcode every path in the program!
A C++ programming language novice, but striving to learn
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Larry Mills Sr wrote: According to Mark CStrings are suppose to be the same as TCHAR's...
What he said was that any place a const TCHAR* is expected, a CString can be used because it has a LPCTSTR() operator.
Larry Mills Sr wrote: ...the code Mark compiled and ran on his machine and he didn't use the _T("%s") or the "L" why won't it on mine?
Maybe his (test) project did not have UNICODE defined.
In any case, if you'll get into the habit of creating Unicode-aware code (e.g., _T(), _tcscpy(), TCHAR), your code will work the same whether UNICODE is defined or not.
"Love people and use things, not love things and use people." - Unknown
"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne
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Larry Mills Sr wrote: the code Mark compiled and ran on his machine and he didn't use the _T("%s") or the "L" why won't it on mine?
Because you changed the code  I used the CString constructor that takes a const char *
as an argument. In a Unicode build, the passed char string is converted to a TCHAR
string (which is wchar_t in a Unicode build).
You changed the code to use an assignment operator. There isn't an CString assignment
operator that takes a const char *, so the code failed to compile.
Larry Mills Sr wrote: But I can't hardcode every path in the program
Of course. But now that you know the file open works with a proper pathname string, you
can work on building a pathname string in a CString. It should be easy to debug - if the
file Open() fails, then you built a bad pathname string
Mark
Mark Salsbery
Microsoft MVP - Visual C++
![Java | [Coffee]](https://codeproject.global.ssl.fastly.net/script/Forums/Images/coffee.gif)
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Larry Mills Sr wrote: CString m_csFileName = ("c:\\Wells\\Program\\DataBases\\Operators\\WSF\\SystemDB.sdb");
There shouldn't be an '=' sign there (at least there wasn't in my code sample).
If you want to use the assignment like that, then as David suggested, you'll
need to assign an appropriate string type, which is a TCHAR:
CString m_csFileName = _T("c:\\Wells\\Program\\DataBases\\Operators\\WSF\\SystemDB.sdb");
Mark
Mark Salsbery
Microsoft MVP - Visual C++
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Oh, it's always the little things. Sorry to have bothered you Mark. I'm so used to using the assignment operator that I failed to notice you didn't. Sorry, and thanks a lot!
A C++ programming language novice, but striving to learn
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Larry Mills Sr wrote: Oh, it's always the little things
And there's so many little things!
Cheers,
Mark
Mark Salsbery
Microsoft MVP - Visual C++
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Another question Mark? Here's how I get the path:
CString CSystemDB::SetPath(CSysData cSysData)
{
CString str, str1, str2;
str1 = "c:\\Wells\\Program\\DataBases\\Operators\\";
// Make path for SystemDB.bku file for each Operator:
pszFileName = _T("c:\\Wells\\Program\\DataBases\\Operators\\WSF\\SystemDB.sdb");// hardcoded
pszFileName2 = _T("c:\\Wells\\Program\\DataBases\\SystemDB\\SystemDB.bku");//hardcoded
// Make path for SystemDB.sdb file for each Operator:
str2 = "c:\\Wells\\Program\\DataBases\\Operator\\";
str2 += cSysData.m_csOperatorName;
str2 += "\\";
str2 += "SystemDB.sdb";
m_csSysNameTMP = str2;
str = str1;
str += cSysData.m_csOperatorName;
str += "\\";
str += cSysData.m_csWellName;
str += "\\";
m_vSys.m_cSysData.m_csPath = str;// ============ Path for file lookup ===================
DoDirectory(m_vSys.m_cSysData);
// Now get the file filename:
cSysData.m_IsFileName = cSysData.m_csWellName;
cSysData.m_IsFileName += ".wdb";
m_csFileName = str;
FileSaveSys(str);// create file =================
return str;
}
You didn't use the assignment operator so how do I get "str" passed to FileSaveSys(str) into the global varable m_csFileName where it will work correctly?
A C++ programming language novice, but striving to learn
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Mark, I don't know how to insert the string that holds the path into the main public CString m_csFileName(?); Here's how I get path CSysData cSysData is just a class for my vector that holds the data. Here's the path code:
Another question Mark? Here's how I get the path:
CString CSystemDB::SetPath(CSysData cSysData)
{
CString str, str1, str2;
str1 = "c:\\Wells\\Program\\DataBases\\Operators\\";
// Make path for SystemDB.bku file for each Operator:
pszFileName = _T("c:\\Wells\\Program\\DataBases\\Operators\\WSF\\SystemDB.sdb");// hardcoded
pszFileName2 = _T("c:\\Wells\\Program\\DataBases\\SystemDB\\SystemDB.bku");//hardcoded
// Make path for SystemDB.sdb file for each Operator:
str2 = "c:\\Wells\\Program\\DataBases\\Operator\\";
str2 += cSysData.m_csOperatorName;
str2 += "\\";
str2 += "SystemDB.sdb";
m_csSysNameTMP = str2;
str = str1;
str += cSysData.m_csOperatorName;
str += "\\";
str += cSysData.m_csWellName;
str += "\\";
m_vSys.m_cSysData.m_csPath = str;// ============ Path for file lookup ===================
DoDirectory(m_vSys.m_cSysData);
// Now get the file filename:
cSysData.m_IsFileName = cSysData.m_csWellName;
cSysData.m_IsFileName += ".wdb";
m_csFileName = str;
FileSaveSys(str);// create file =================
return str;
}
You didn't use the assignment operator so how do I get "str" passed to FileSaveSys(str) into the global varable m_csFileName where it will work correctly?
A C++ programming language novice, but striving to learn
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I didn't use the assignment operator, but I didn't say there was
anything wrong with using it. You just need to assign a type that's
compatible with one of the assignment operators for the CString class.
You have a CString you're passing to FileSaveSys(), but you also already assign
it to a variable called m_csFileName in your SetPath() method.
I'm not sure why you pass it AND set a member variable...either way, you can use it
in FileSaveSys() however you want - use the passed string or the member variable string...
void CSystemDB::FileSaveSys(CString str)
{
CStdioFile myFile;
CFileException ex;
if(!myFile.Open(<code>m_csFileName</code>, CFile::modeNoTruncate | CFile::modeCreate | CFile::modeReadWrite | CFile::shareDenyWrite | CFile::typeText, &ex))
{
...
void CSystemDB::FileSaveSys(CString <code>str</code>)
{
CStdioFile myFile;
CFileException ex;
if(!myFile.Open(<code>str</code>, CFile::modeNoTruncate | CFile::modeCreate | CFile::modeReadWrite | CFile::shareDenyWrite | CFile::typeText, &ex))
{
...
Mark Salsbery
Microsoft MVP - Visual C++
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This is why:
C:\Wells\Program\DataBases\Operator\WSF\SystemDB.sdb contains an incorrect path. (error message)
"C:\Wells\Program\DataBases\Operator\WSF\SystemDB.sdb" is what is in m_csFileName and the path is correct.
if I hardcode it,(exactly as above), in the constructor it does NOT produce this error. I've tried it so I know this is true.
A C++ programming language novice, but striving to learn
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Mark, I want to appolize to you for taking up so much of your time. I finally found my mistake and it was as you said THE PATH WAS WRONG! It was writing all along but I diden't know it; because it was putting it somewhere else.
Sorry for using your time. I tried to look it up in debug but over looked it!
A C++ programming language novice, but striving to learn
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Good deal! Glad you found it
Cheers,
Mark
Mark Salsbery
Microsoft MVP - Visual C++
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Hi all,
I want to know how can i use keyboards kyes like enter,move-down/up/left/right.
Please tell me how can use Enter key in Tree view.and how can select a tree item with the help of Enter key.
thanks in advance.
IN A DAY, WHEN YOU DON'T COME ACROSS ANY PROBLEMS - YOU CAN BE SURE THAT YOU ARE TRAVELLING IN A WRONG PATH
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About the Keys, take a look into this message[^] and its answers. With the CTreeCtrl, sorry but I haven't used it yet. Have you checked in the articles? I remember seeing one about an extended tree control, maybe there is info you can use.
Greetings.
--------
M.D.V. 
If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about?
Help me to understand what I'm saying, and I'll explain it better to you
“The First Rule of Program Optimization: Don't do it. The Second Rule of Program Optimization (for experts only!): Don't do it yet.” - Michael A. Jackson
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ho can replace a view on click of enter key?
IN A DAY, WHEN YOU DON'T COME ACROSS ANY PROBLEMS - YOU CAN BE SURE THAT YOU ARE TRAVELLING IN A WRONG PATH
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I find that in my Visual C++ 2008 I can select between Debug and Release modes in two places:
A: at the top of the window just below the Help menu option.
B: reachable by clicking Build, Configuration Manager.
I have found that my C:\3d\makeobj_vc\... keeps two copies of its working: Debug and Release.
What happens with each of these four selections for A and B?:-
A and B both Debug
A and B both Release
A Debug, B Release
A Release, B Debug
modified on Saturday, May 3, 2008 7:23 AM
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Didnt get you completly what you wanna say.
When i do changes from A than automatically B also changes.
It is not possible that if one changes than another is not change.
3) A Debug, B Release
4) A Release, B Debug
please check it again. I think you do changes in project context (configuration manager)not active solution configuration. If yes than when the both condition are different than project is not built and failure occur.
Yes U Can ...If U Can ,Dream it , U can do it ...ICAN
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hi Everyone
i am doing project on GSM modem using AT commands in VC++ programming.
I am not getting how to send,read and receive a sms from GSM modem using AT commands.I am opening port and connecting modem.it is working.But i am not getting how the message shold me send to particular
persons in the list.whether i have to write a function for AT commands or wt.if it is so then tell me how can i do it.
Thanks,
Savitri
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savitri wrote: i am doing project on GSM modem using AT commands in VC++ programming.
I am not getting how to send,read and receive a sms from GSM modem using AT commands.I am opening port and connecting modem.it is working.But i am not getting how the message shold me send to particular
persons in the list.whether i have to write a function for AT commands or wt.if it is so then tell me how can i do it.
do you have manual for AT command if yes. then try connecting hardware usin hyper terminal and there you could practice all of your command!
as far a i remember, i used to practice these command on hyperterminal
ctrl + b SMAN 0 ctrl+c
"Opinions are neither right nor wrong. I cannot change your opinion. I can, however, change what influences your opinion." - David Crow
Never mind - my own stupidity is the source of every "problem" - Mixture
cheers,
Alok Gupta
VC Forum Q&A :- I/ IV
Support CRY- Child Relief and You/codeProject$$>
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Hi..
Thakns for replying.i tried in hyperterminal but i am not getting how to use those commands in my program.when i open the port it is working.but i want to use AT+CMGS AND AT+CMGR commands in my program.is there any syntax for it.please tell me i am new to this AT commands.PLEASE HELP ME OUT.
PLEASE........
THANKS IN ADVANCE
Savitri
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Hello everyone,
My question is comes from when I write a C++ program. Not find suitable place to post.
My question comes from page 175 of Numerical Computation Guide. Here is a link,
http://www.physics.ohio-state.edu/~dws/grouplinks/floating_point_math.pdf
I do not know why if ß = 2, p = 3, emin = -1 and emax = 2 there are
16 distinct floating-point values in formular (1) in page 174? Are
there any more general formular to calculate the distinct value for
formular (1) for any given ß, p, emin and emax?
(sorry, I can not post math formular in this text box)
thanks in advance,
George
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ß is the base we're working in. When ß is 2, we're dealing with binary floating point.
If emin is -1 and emax is 2, there are four possible values for e: -1, 0, 1, and 2. p represents the number of significant binary digits, but because we're talking about normalized floating point numbers, d0 cannot be 0. Because we're dealing in binary, it must be 1. Therefore, only d1 and d2 can change. Because there are two bits, that means there are four possibilities (1.00, 1.01, 1.10, 1.11). Combining the four possible values of the mantissa with the four possible exponents gives 16 possibilities:
1.00 x 2<sup>-1</sup> = 0.1 (== ½, 0.5 decimal)
1.01 x 2<sup>-1</sup> = 0.101 (0.625)
1.10 x 2<sup>-1</sup> = 0.11 (0.75)
1.11 x 2<sup>-1</sup> = 0.111 (0.875)
1.00 x 2<sup>0</sup> = 1.0 (1.0)
1.01 x 2<sup>0</sup> = 1.01 (1.25)
1.10 x 2<sup>0</sup> = 1.10 (1.5)
1.11 x 2<sup>0</sup> = 1.11 (1.75)
1.00 x 2<sup>1</sup> = 10.0 (2.0)
1.01 x 2<sup>1</sup> = 10.1 (2.5)
1.10 x 2<sup>1</sup> = 11.0 (3.0)
1.11 x 2<sup>1</sup> = 11.1 (3.5)
1.00 x 2<sup>2</sup> = 100.0 (4.0)
1.01 x 2<sup>2</sup> = 101.0 (5.0)
1.10 x 2<sup>2</sup> = 110.0 (6.0)
1.11 x 2<sup>2</sup> = 111.0 (7.0) [Edit: fixed formatting for clarity]
DoEvents: Generating unexpected recursion since 1991
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Cool, Mike! Smart guy!
1.
What is the range of the digit in the left of the point in normalized form of float, in base B? [1, B - 1]?
2.
Your conclusion is matching the answer from the book.
But what makes me confused is, suppose you have significant p1 and p2, and the correspounding exponent e1 and e2. Your counting answer is based on the theory that if p1 and p2 are different, no matter what correspounding values are e1 and e2 are, the result must be different.
My question is, is it possible that p1 and p2 are different, but also e1 and e2 are different, then p1 * e1 and p2 * e2 produce the same resulting value?
regards,
George
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