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This was the answer I was looking for, thank you.
Regards,
Mike
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mla154 wrote: This was the answer I was looking for
but it wasn't full anyway.
I maintain that Hamid's solution, such as the menu item you found will only work in the Resource ("Layout") Editor...
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I appreciate everyone's responses anyway. I understand now that the user must have the Resource Editor open and then press Ctrl-D.
Regards,
Mike
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you have to be in the "layout editor" to have it enabled
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mla154 wrote:
I have created a dialog using Visual C++ 2005. On the dialog are 2 group boxes and 2 ActiveX controls. How do I change the tab ordering of these objects? I have found the Layout toolbar, which contains the Tab Order button, but I don't know how to enable this button. Any help is appreciated.
This works in the resource editor.
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Hello,
If I have the following code,
<br />
#include <iostream> <br />
<br />
int main()<br />
{<br />
int y = 6;<br />
int *ptr;<br />
ptr = &y;<br />
<br />
int x = 7;<br />
<br />
ptr = &x;<br />
<br />
x = 8;<br />
std::cout << "hi";<br />
std::cout << "hello";<br />
return 0;<br />
}<br />
is there a way to somehow change the break point properties so that it will stop where ptr changes from &y to &x ? The reason I wish to know this is lets say I have a massive call stack and somewhere in there my pointer is modified, I wish to locate this position in the stack then by setting a break point where the properties of my pointer change.
The platform I am using is visual studio .NET. I believe it is possible to do this in visual studion 6 C++ but I am unsure how to do this in visual studio .NET.
Thanks for any input.
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yes, place a "conditional breakpoint" instead.
on that breapoint, check it's breaking condition, and add the values to test...
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Hi all,
I m using a bmp at uper side of dialog , i m placing a static text over the .bmp.
but when i run the application the text is hide.
same problem occur in case of button.
please tell me what can i do.
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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Does this help?
"Love people and use things, not love things and use people." - Unknown
"The brick walls are there for a reason...to stop the people who don't want it badly enough." - Randy Pausch
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thanks sir.
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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Hamid. wrote: And I think Displaying split Bitmap files[^] is helpful for you.
How?
"Love people and use things, not love things and use people." - Unknown
"The brick walls are there for a reason...to stop the people who don't want it badly enough." - Randy Pausch
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I saw your answer was perfect so I guess maybe this article is helpfuls for "_$h@nky_" on the othe aspects.
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that's nice article. who downvoted you.. let me clear the air!
"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/xml>
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Enemy!
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let's kill them then
"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/xml>
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Hi all,
i have done little calculation.
ULONGLONG test;
long double final;
test=test+1024;
final=test/1048576;
here if the actual value of final thru debugging is 7.260000, but it show only 7.00000000;
please tell me how can i get exact value
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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It looks to me (though I could be wrong) like this is a problem with an implicit cast. What the compiler is doing is this :
final = (long double)(test/1048576);
but what you actually want it to do is this :
final = ((long double)test)/1048576;
Give it a try. Hope that helps.
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That won't solve the problem neither: you are still dividing by an integer (1048576), so the result will be an integer.
EDIT: oooops, I don't know what I smoke this morning but I am completely wrong. Once one of the operand is a double, the result will be a double too. The problem occurs once both operands are integers, then the result will be an integer too. So, casting one operand to a double as you did correct the problem.
modified on Monday, August 18, 2008 8:19 AM
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The problem is because you are dividing by an integer. Thus, the compiler will implicitely cast your double into an integer in order to make an integer division (thus, the result will be an integer).Try this:
test = test + 1024.0;
final = test/1048576.0;
EDIT: as I explained here[^], I was not very awake this morning . So, once you have at least one double in the operands, then the result will be a double too. If both operands are integers, then the result is an integer too.
modified on Monday, August 18, 2008 8:23 AM
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test is an integer so there's no benefit to adding a floating point
const to it.
What's important is coercing the integers to doubles before performing
the division
Mark
Mark Salsbery
Microsoft MVP - Visual C++
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Mark Salsbery wrote: test is an integer so there's no benefit to adding a floating point
const to it.
Yes sure. But if you see my other post, you can see that I was wondering what I smoked this morning
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I wonder that often myself (what you smoked this morning)
(kidding, of course)
Mark Salsbery
Microsoft MVP - Visual C++
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C language tutorial needed.
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
[Image resize DLL]
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