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your maen to say that I can use GDI+ 1.0 on windows viats and windiws 7, correct ?
I am developing application on WIndows XP SP3 and I checked gdiplus verison that us 5.1.3
But you mentioned only 2 type of gdi+.
I am confused, please make me clear :
1. Is it manadetory to give gdi+ with application to run on any version of window (above XP), if application is developed on windows XP (SP3).
2. Which version need to give with application?
3. I did not get any redistribute issue with gdi+, please give link?
Thanks,
JM
modified on Wednesday, August 18, 2010 10:12 AM
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john5632 wrote: your maen to say that I can use GDI+ 1.0 on windows viats and windiws 7, correct ?
Yes. Windows 7 even ships with both versions of GDI+.
If your program is GDI+ 1.0 based then it should work for later versions of GDI+,
so it is normal to work on Windows Vista and Windows 7.
But in case of problems you can redistribute GDI+ 1.0 to Windows Vista.
This[^] is a link for redistributable package for GDI+ 1.0.
john5632 wrote: I checked gdiplus verison that us 5.1.3
This is a file version of the DLL file, not the version of the GDI+ library.
john5632 wrote: 1. Is it manadetory to give gdi+ with application to run on any version of window (above XP), if application is developed on windows XP (SP3).
Yes. GDI+ 1.0 can be redistributed, I've already provided a link for the redistributable package.
john5632 wrote: 2. Which version need to give with application?
You can only give GDI+ 1.0, this version is used by Windows XP. GDI+ 1.1 is not redistributable according to EULA of Windows Vista.
john5632 wrote: 3. I did not get any redistribute issue with gdi+, please give link?
Once again here[^] is the MSDN link.
Also for additional details see the documanetation[^] of GDI+ (this link was already given to you by signor Pallini).
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Nuri Ismail wrote: john5632 wrote:
1. Is it manadetory to give gdi+ with application to run on any version of window (above XP), if application is developed on windows XP (SP3).
Actually no, it is NOT mandatory, in fact there may be no reason to redistribute GDI+.
Did you develop and test your application on a Windows XP system? If so, and you have not tried to force the 1.1 issue, then your application is limited to GDI+ 1.0 functionality. Therefore, you can redistribute your application to Windows XP, Vista, and Windows 7 systems without needing to redistribute the GDIplus.dll. That dll is already on those systems. On the other hand, do you want to distribute your application to downlevel systems like Windows 98? If so, then you provide the redistributable 1.0 version of GDIplus.dll for those downlevel computers and only them, not for ones where gdiplus.dll is provided with Windows.
Do you want to use some of the advanced functionality in GDIplus 1.1, such as effects? If so, you must define switches so that the added 1.1 functionality will be available when you compile. You must also add a manifest so that gdiplus.dll version 1.1 will be loaded when your programme executes. You can use this version of your programme on XP, Vista, and Windows 7. Again, you do not provide the dll, it is part of Windows.
Finally, in Windows 7 and beyond, GDIplus.dll version 1.1 is the default version used. This means that your programme, developed under and for Windows XP, without a version 1.1 manifest, will load and use the GDIplus.dll version 1.0. Under Vista, the 1.0 version will also be used. But, on Windows 7, GDIplus.dll version 1.1 will be loaded by that same application. As it is a full superset, you should not see any differences.
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Here [^] some info:
Run-time Requirements
Gdiplus.dll is included with Windows XP. For information about which operating systems are required to use a particular class or method, see the More Information section of the documentation for the class or method. GDI+ is available as a redistributable for Microsoft Windows NT 4.0 SP6, Windows 2000, Windows 98, and Windows Millennium Edition (Windows Me). To download the latest redistributable, see [^].
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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Hello All,
I made an API Hook using mHook Library and successfully hooking NtSetInformationFile , I then made an DLL which loads the DLL into the process using CreateRemoteThread, It's working fine in my PC, But when checked on windows server 2003(32 bit) the application won't start and gives me the message that "This application has failed to start because the application configuration is incorrect. Reinstalling the application may fix the problem." .
Now i googled about the error and found so many answers, One of them was to check the "Code generation" setting under C/C++ property, if it is Multi-Threaded DLL (/MD) then change to Multi-Threaded (/MT) , Changing the setting works in windows server 2003, But now the problem is when i recompile my code with (/MT) the anti virus won't let me build the exe and delete the exe, I can only build and run the exe after i uninstall or stop the anti-virus. Could anyone throw some light on this problem.
Thanks
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Do not change the property to multithreaded.
Please follow the steps:
1) Go to Project Menu -> Properties
2) Select Configuration Properties ->general
There are two properties
Use of MFC :if you have "Use MFC in a Shared DLL" than change this property to "Use MFC in a Static Library".
Use of ATL :if you have "Dynamic Link to ATL" than change this property to "Static Link to ATL".
than check.
Yes U Can ...If U Can ,Dream it , U can do it ...ICAN
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The value under "Use of ATL" is "Not Using ATL"
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If your setting is "Not using ATL" than dont change the setting if your setting is "Dynamic Link to ATL" than change it to "Static link to ATL" and what about MFC setting.
Yes U Can ...If U Can ,Dream it , U can do it ...ICAN
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MFC Setting is "Use Standard Windows Libraries"
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I am not sure please change the settings of "Use Standard Windows Library" to "Use MFC in a Static Library" and compile.
if u receive any error of start up than update the setting of Linker Entry point and than check if it is working fine.
Yes U Can ...If U Can ,Dream it , U can do it ...ICAN
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I added msvcrt.lib and it's working
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Thats Gr8
Yes U Can ...If U Can ,Dream it , U can do it ...ICAN
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You, sir, are tied for having the world's lamest signature.
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Hi All,
In my website there is facility that user can send mail using it own email SMTP setting. some user is sending mail success fully but some facing problem for this there is option in config page in which user have to enter his SMTP setting. And also there is test button when users press a test button and if any wrong entry (username, password, smtpServer) then I want to show pop message accordingly.
Is any way in which we can identified that what is wrong entry (username, password or mail server) because I want to give message like "wrong username or wrong SMTP server.
Thanks
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Samarjeet Singh@india wrote: Is any way in which we can identified that what is wrong
You could start by posting your question in the correct forum.
It's time for a new signature.
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Realy I am so sorry.
is there any way by which we can change this post in correct(asp.net)forum?
thanks
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Don't worry about this as it has now been marked. Just repost in the ASP.NET forum and hope you get a good answer.
It's time for a new signature.
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Hi sir,
I have origin axis of bitmap and x1,x2,y1,y2.
I want to draw a rectangle and rotate it with the given degree.I am trying with this code,but i am not getting.
Can some one help me
double x1,x2,y1,y2;
sf=(1024/31);
A1 = 10.00;
A2 = 10.00;
B1 = 10.00;
B2 = 10.00;
x1 = (A1 * sf);
x2 = (A2 * sf);
y1 = (B1 * sf);
y2 = (B2 * sf);
center.x=(((pCellInfo->rcBitmapRect.right-pCellInfo->rcBitmapRect.left)/2)+pCellInfo->rcBitmapRect.left);
center.y=(((pCellInfo->rcBitmapRect.bottom-pCellInfo->rcBitmapRect.top)/2)+pCellInfo->rcBitmapRect.top);
int a = 45;
float Angle = ( 3.142 * a ) / 180;
x1 = x1 * cos(Angle) + y1 * sin(Angle);
y1 = -x1 * sin(Angle) + y1 * cos(Angle);
x2 = x2 * cos(Angle) + y2 * sin(Angle);
y2 = -x2 * sin(Angle) + y2 * cos(Angle);
Origin(525,454);
MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL);
LineTo(pCellInfo->hDC,center.x-x1,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y-y2);
LineTo(pCellInfo->hDC,center.x-x1,center.y-y2);
If i test with this hard code values,i am getting the rectangle rotated
MoveToEx(pCellInfo->hDC,525,289,NULL);
LineTo(pCellInfo->hDC,360,454);
LineTo(pCellInfo->hDC,525,619);
LineTo(pCellInfo->hDC,690,454);
LineTo(pCellInfo->hDC,525,289);
how can i get the same values
Thanks
Raj
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raju_shiva wrote: i am not getting
What does it mean, exactly (i.e. please elaborate)?
raju_shiva wrote: A1 = 10.00; //these are values i am reading from hardware
A2 = 10.00;
B1 = 10.00;
B2 = 10.00;
How do you hope to get a rectangle from these values?
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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CPallini wrote: . please elaborate)?
These are left,top,right,bottam values for the rectangle
i.e Rectangle(__in HDC hdc, __in int left, __in int top, __in int right, __in int bottom);
Thanks
Raj
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{10,10,10,10} as {left,top,right,bottom} values give an empty rectangle.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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CPallini wrote: {10,10,10,10} as {left,top,right,bottom} values give an empty rectangle
I am not passing the values directly.
As it is a bitmap image,on that i am drawing the rectangle
I am multiplying it :
x1 = (A1 * sf);
x2 = (A2 * sf);
y1 = (B1 * sf);
y2 = (B2 * sf);
where sf is
sf=(1024/31); //Its a bitmap image
Then while drawing the rectangle,i start as (center.x-left)
MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL);
I hope you got it.
Thanks
Raj
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