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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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raju_shiva wrote: x1 = (A1 * sf); //i.e (10 * 33.03)
x2 = (A2 * sf); //i.e (10 * 33.03)
y1 = (B1 * sf); //i.e (10 * 33.03)
y2 = (B2 * sf); //i.e (10 * 33.03)
I see no rectangle here.
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: I see no rectangle here
After getting all the values
Here i am drawing the rectangle
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);
Thanks
Raj
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OK (sorry if I didn't get you).
Now, what is the problem with your code (expected behaviour vs observed one)?.
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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Now i want to rotate it for the given degree.How can i do it.
I am bit confused
please help me
Thanks
Raj
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You are defining the rectangle via offsets from center , that is
P0={-x1,-y2}, P1{-x1, y1}, P2={x2,y1}, P3={x2,-y2}
hence, if you wan't rotate with angle phi around the center , than you should compute:
Pr = { x * cos(phi) + y * sin(phi), -x * sin(phi) + y * cos(phi)}
i.e.:
P0R= { -x1 * cos(phi) -y2*sin(phi), x1 * sin(phi) - y2 * cos(phi)}
P1R =...
P2R =...
P3R =...
and then connect the center+PiR points the way you did before.
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: P0R= { -x1 * cos(phi) -y2*sin(phi), x1 * sin(phi) - y2 * cos(phi)}
I am confused with it???
Now suppose my values are:
x1 = 330,x2 = 330,y1=330,y2=330
Origin(center.x,cebter.y) = Origin(525,454);
I have to do the get the new x1,x2,y1,y2 before calling MoveToEx and LineTo
Am i right??
i.e
Pr = { x * cos(phi) + y * sin(phi), -x * sin(phi) + y * cos(phi)}
for each x1,x2,y1,y2.
Then call the function
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);
Thanks
Raj
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Something like this
double phi = atan(1.0) * 2/3; int x[4];
int y[4];
x[0] = -x1 * cos(phi) - y2 * sin(phi);
y[0] = x1 * sin(phi) - y2 * cos(phi);
x[1] = -x1 * cos(phi) + y1 * sin(phi);
y[1] = x1 * sin(phi) + y1 * cos(phi);
x[2] = x2 * cos(phi) + y1 * sin(phi);
y[2] = -x2 * sin(phi) + y1 * cos(phi);
x[3] = x2 * cos(phi) - y2 * sin(phi);
y[3] = -x2 * sin(phi) - y2 * cos(phi);
for (int i=0; i<4; i++)
{
x[i] += center.x;
y[i] += center.y;
}
MoveToEx(pCellInfo->hDC,x[3],y[3],NULL);
for (int i=0; i<4; i++)
{
LineTo(pCellInfo->hDC, x[i],y[i]);
}
I suppose.
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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Thanks a lot ,its working fine.
Thanks
Raj
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You are welcome.
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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The code looks correct but I find it to be slightly offensive to me sensibilities.
I'm just kidding. I usually make a function or method when I see a sequence of code repeated more than twice. Something like this is generic enough that I definitely would.
Of course, it's not your job to put this is into a function. That is an exercise for the reader.
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