|
Is your editbox common control?
Set dialog's language property to japanese that dialog has the
editbox.
|
|
|
|
|
Thank for your reply:
>Is your editbox common control?
Yes, it's.
>Set dialog's language property to japanese that dialog has the editbox.
I seted dialog's language property to Japanese. But it still show message by English.
Is there any function to set again language?
|
|
|
|
|
if you are static linking MFC, try changing it to use shared MFC DLL
(in project settings)
|
|
|
|
|
I changed it to use shared MFC DLL. But i get some error:
error LNK2001: unresolved external symbol ___argc
error LNK2001: unresolved external symbol ___argv
error LNK2001: unresolved external symbol __mbctype
error LNK2001: unresolved external symbol __mbctype
error LNK2019: unresolved external symbol __mbctype referenced in function "void __stdcall _AfxAbbreviateName(char *,int,int)" (?_AfxAbbreviateName@@YGXPADHH@Z)
fatal error LNK1120: 3 unresolved externals
|
|
|
|
|
I'm sorry, I can't think of what is causing these errors
if this is part of a large project, I suggest you first try what you want to do with a simpler project containing only a single edit box, with range limiting
I suppose you are building on an english OS and then copying the binary to a japanese OS. the problem may be that you are statically linking the library that produces the message, so changing to the shared DLL should make the program use the library installed on the OS (you obviously need to install MFC on the target machine)
hope this helps...
I'm unable to reproduce your problem since I've only got one OS (japanese)
|
|
|
|
|
|
and...
Set >Main< Dialog's language property to janpanese.
else, remake dialog. It may be that bug.
^^;
|
|
|
|
|
use UpdateData() at the last after you programatically check ,if its not in the range provide your on messagebox which takes string from the japanese Resource by this way you can generealize your application.
Hope this solves your problem
Regards
Farpointer
-- modified at 11:29 Saturday 22nd October, 2005
|
|
|
|
|
How to create new thread in VC??
Hung
|
|
|
|
|
If using MFC, AfxCreateThread()
If using just CRT, _beginthreadex()
Anyone who thinks he has a better idea of what's good for people than people do is a swine.
- P.J. O'Rourke
|
|
|
|
|
|
How to show Balloon tips!(not tooptip)
We can see this when connected ext storage drives.
Do not VS6.0(with Platform SERVER 2003 feb SDK) support?
-- example --
m_nidIconData.cbSize=sizeof(NOTIFYICONDATA);
m_nidIconData.uFlags = NIF_INFO;
m_nidIconData.uTimeout = uTimeout;
m_nidIconData.dwInfoFlags = dwInfoFlags;
strcpy(m_nidIconData.szInfo, szMsg ? szMsg : _T(""));
strcpy(m_nidIconData.szInfoTitle, szTitle ? szTitle : _T(""));
Shell_NotifyIcon( NIM_MODIFY, &m_nidIconData );
-- error result --
'NIF_INFO' : undeclared identifier
Answer plz~ ^^
|
|
|
|
|
I think it's NIF_MESSAGE instead of NIF_INFO
SkyWalker
|
|
|
|
|
|
ch2ho wrote: -- error result --
'NIF_INFO' : undeclared identifier
Check make sure that VS6 is using the Platform SDK includes.
DEBUGGING : Removing the needles from the haystack.
|
|
|
|
|
I wrote "with Platform SERVER 2003 feb SDK" in prepost.
I had checked.
Anyway, thank you~.
|
|
|
|
|
Hi,
I created an EDITBOX.
I want it limited up to 5 upper case characters.
I set the property upper case to True. So, it works
fine with that but I am not sure how to limit
the user to enter up 5 characters.
Any help would be appreciated!
Thanks,
Kevin
|
|
|
|
|
If using MFC, look at CEdit::SetLimitText() call.
If using plain windows, look at the EM_LIMITTEXT message.
Anyone who thinks he has a better idea of what's good for people than people do is a swine.
- P.J. O'Rourke
-- modified at 21:55 Friday 21st October, 2005
|
|
|
|
|
Hi friends,
I have developed one application and i want to measure the time take by each module. I can get the time difference if it is in second however few module took time less than second. Can you please help me how to get the difference when it is millisecond or less than that. Thank you in advance.
Regards
-- modified at 17:30 Friday 21st October, 2005
|
|
|
|
|
Hi,
// Get the current time (start time).
DWORD dwStartTime = GetTickCount();
// The stuff here takes some time
// Subtract start time from current time to get duration.
DWORD dwElapsedTime = GetTickCount() - dwStartTime;
/////////////////////
Also see the APIs:
GetThreadTimes
GetProcessTimes
--
======
Arman
|
|
|
|
|
Are you sure it will give the time difference less than second?
Thanks
|
|
|
|
|
for higher accuracy than this method, look up high resolution performance counters here
link: MSDN - Timers[^]
|
|
|
|
|
Yes it does. It gives the time difference in milliseconds.
--
======
Arman
|
|
|
|
|
 I am using Visual C++ 6.0 MFC with a single document application. I can create a graphic of a wafer map showing the failures by bin number. There are 32 different bin numbers. I can bring up a dialog that alows me to select the bin numbers to display and then print that current view. But I need to print the wafer map for each bin, which means that I have to open the dialog box and check bin 1 and then print, then open the dialog box and check bin 2 and print and so on 32 times. What I want to do is simply tell the application to print by bin and have 32 maps printed, one for each bin number. How would I go about doing this?
Buck
|
|
|
|
|
I have this code that doesn't work. I try to convert an image to the frequency domain using dct and then back
numx: width in bytes of the image
numy: height
ppixel[]: the pixels of the image. I use them as a 1-D array.
The code goes like this:
<code>
/********************************Starting the DCT transform**/
double pi = 3.14159265359;
double sum=0;
double *pdct=new double[numx*numy]; //the dct coefficients
for (int xx=0;xx<numx*numy;xx++)
{
sum+=ppixel[xx];
}
pdct[0]=sqrt(1/(numx*numy))*sum; //dct for u=0 (DC)
sum=0; //resetting
for (int u=1;u<numx*numy;u++)
{
for (int x=0;x<numx*numy;x++)
{
sum+=ppixel[x]*cos((pi*(2*x+1)*u)/(2*numx*numy));
}
pdct[u]=sqrt(2/(numx*numy))*sum; //the dct for u!=0 (AC)
}
</code>
My idct code is:
<code>
/********************************Creating the IDCT transform**/
sum=0; //resetting
for (int x=0;x<numx*numy;x++
{
for (int u=1;u<numx*numy;u++)
{
sum+=sqrt(2/(numx*numy))*pdct[u]*cos((pi*(2*x+1)*u)/(2*numx*numy));
}
ppixel[x]=sqrt(1/(numx*numy))*pdct[0]+sum; //the first is for u=0 and the sum for u!=0
}
</code>
Is there somwthing wrong that I don't see? Can anyone help me?
Thanks..
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
