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Hi.
I want to use some .Net components from System::Xml and System::Xml::Xsl. But I can't seem to access these from my MFC application. I have set up the compiler options, and other options to allow use of .Net components, but when I write the code I dont manage to access these objects... Can anyone tell me what I have to do in order to do this. Any tutorials / articles on the subject are highly appreciated too
Thanks.
J.
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how do i get around this?
using pow() from math.h:
-1 raised to a fraction power should return undefined (-1.#INF or whatever) when the
denominator is even, yet it should return -1 when the denominator is odd
e.g. -1 ^ 1/3 = -1, while -1 ^ 1/2 = undef
so how can i get it to return -1 instead of undef for all fractional powers?
r -€
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Which are you asking about? How to change the behavior of pow() with an even denominator, or the return value of pow(-1, 0.3333333333) not being what you expect?
For the former, you'll need to write your own wrapper around pow() since you want to change the behavior of pow().
For the latter, 0.3333333333 is not 1/3 so I don't know how pow() will handle it. The answer to that probably lies in the depths of the floating point number representation, which I don't know enough about to give a good answer.
--Mike--
Personal stuff:: Ericahist | Homepage
Shareware stuff:: 1ClickPicGrabber | RightClick-Encrypt
CP stuff:: CP SearchBar v2.0.2 | C++ Forum FAQ
----
Actual sign at the laundromat I go to: "No tinting or dying."
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thanks... my problem is that im making a calculator program and when someone types "(-1)^(1/3)" it returns undefined... which is wrong, the answer is 1
i guess i can't get around that issue unless i represent fractions as numerator/denominator instead of the actual ratio value
oh well
thanks for the help though
r -€
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Um, no. (-1)^(1/3) = 0.5+0.866i (as one of many roots. -1 is another root).
"pow" only supports real numbers, not imaginary.
FYI: 1^3 != -1, thus your answer is obviously in error.
Tim Smith
I'm going to patent thought. I have yet to see any prior art.
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well, the cube root of -1 is -1 (right?) and a number to a fraction power (a/b) is the b'th root of the funtion to the a'th power... am i not correct?
r -€
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Is "the" cube root of -1, -1?
No.
It is one of at least two cube roots. That is part of problem. Probably the real problem is that you can probably only compute a root when 1/x and x is an integer. As someone already pointed out, due to floating point limitations, it is next to impossible to get 1/x and have x an integer.
Tim Smith
I'm going to patent thought. I have yet to see any prior art.
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>Is "the" cube root of -1, -1?
>No.
>It is one of at least two cube roots.
By that logic, pow(1, 1/3) should return undefined too, since -1/2 + j * sqrt(3)/2 and -1/2 - j * sqrt(3)/2 are also solutions. (It doesn't. It returns 1, as you would expect.)
For what it's worth, the calculator in Windows XP returns "Invalid input for function" when you try to do (-1)^(1/3). I think that's what most calculators that don't handle complex numbers do.
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Read the second half of the message please. I amended my response long before you posted your message where I talk about the problem of detecting the 1/x where x is an integer.
Tim Smith
I'm going to patent thought. I have yet to see any prior art.
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Sorry. I didn't mean to offend you. I guess I should have worded that more carefully.
I was just trying to point out that his question was a little deeper than people were giving it credit for. As you said, the floating point representation of 1/3 is not exact. Yet, pow(1,1/3) correctly returns 1, even though there are two equally valid complex solutions. Why, then, shouldn't pow(-1,1/3) return -1? It is, after all, the only real-valued solution. I guess the ANSI C developers opted for efficiency rather than completeness.
Again, sorry if I offended you. That wasn't my intention.
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Hi there
Tim is absolutely right.
The easiest way to calculate a power is using logarithms, so let say we have this
r = x ^ y
so to calculate r will be something like this
r = e ^ (y * log x)
where e^ and log can be obtained using the Taylor Polynomial
f(x) = f(a) + f'(a) * (x-a)/1! + f"(a) * (x-a)^2/2! + ...+ fN(a) * (x-a)^n/n!
where f', f" ... fN are the first n derivatives of f, and a is a value with which all the funcion can be calculated, for e^ can be 0 for ln, 1.
Anyway, like I was saing the use of logarithm is the reason you are not geting the real root you are looking for, as you can see in your case you would have something like this
r = e ^ ( 1/3 * log -1)
and as we all know log -1 has no solution on the domain of real numbers, the solution is an imaginary number, -PI i, and there is your problem.
So you have some options here, one is to do some analitical preprocessing before the actual calculations, the other one is to redefine pow and/or log to handle complex numbers.
Fabian
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I think it is because of this reason:
a^x = y
ln(a^x) = ln(y)
x.ln(a) = ln(y)
e^(x.ln(a)) = y
ln(a) is undefined for a <= 0
it should work for a,x,y are of type complex. I have done such a thing many years ago. It works also for -1 ^ 1/2 (which is the complex number "i" or "j") written as y.re=0 and y.im=1 .
I am not sure if pow can be called as "complex y = pow( complex a, complex x )", if not, you need to overload it in some way. That's a challenge!!
Wish it helps some.
Michel
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thanks, i get what you're saying
i think maybe i will implement sets and complex numbers into my calculator program to get around this..
i've never heard of the "complex" class, though.. is it STL?
so far i'm only using STL classes and would like to keep it to a bare minimum of external libraries
r -€
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Hi,
Yes Complex is a class of STL, but it won't help you either, try this
complex<double> x, y, r;
x.real(-1.0);
x.imag(0.0);<br> <br>
y.real(1.0/3.0);
y.imag(0.0);<br> <br>
r = pow(x, y);<br>
r = exp(y * log(x));
So as I said, you will have to do some preprocessing or redefine pow yourself
Fabian
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thanks for the info... ill leave the fix for a later date
r -€
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I use Visual C++ 6.0. I want to link my MFC application staticly instead of using shared DLLs. If I go to Project->Settings->General I can ONLY chose "Use MFC in a shared DLL". What's wrong?
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::VkKeyScan translates a character to the corresponding virtual-key code and shift state for the current keyboard, but which function does the reverse and translates the VK code and shift state to a character, is there any? Thanks.
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ToAsciiEx / ToUnicodeEx.
Stability. What an interesting concept. -- Chris Maunder
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i want to make a keyboard driver. my aim is to type any Indian languages in ms word or excel or notepad etc.what are the code words for hooking ( trap ) a key stroke and post it to a desired application. i have some idea first we have to get the focus of the window ( get focus api) then each keystroke has to be trapped and converted to their ascii. Then using post message api converted ascii is passed to the input focus window.
but i am not sure, pls give me a small program by which when we type 'a' in the keyboard we must get 'b' in the desired window. thanks
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I think you will need to install a keyboard hook. Check ::SetWindowsHookEx.
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I'm trying to display a small icon in a CStatusBarCtrl. I created a 16x16 icon, IDI_X, in the resource editor, then I tried the following:
HICON hIcon;
hIcon = theApp.LoadIcon(MAKEINTRESOURCE(IDI_X));
m_wndSBC.SetIcon(0, hIcon);
The icon is resized to 32x32 and doesn't fit in the status bar. The CApp::LoadIcon documentation says it can only be used to load an icon whose size conforms to the SM_CXICON and SM_CYICON system metric. It says to use LoadImage instead. So, I tried this:
HANDLE hIcon;
hIcon = LoadImage(NULL, MAKEINTRESOURCE(IDI_X),
IMAGE_ICON, 0, 0, 0);
DWORD dwError;
dwError = GetLastError();
That didn't work either. LoadImage returns NULL and GetLastError returns 1813: "The specified resource type cannot be found in the image file."
Can anyone help?
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Just an update. I got it to work by doing the following:
HANDLE hIcon;
hIcon = ::LoadImage(NULL, "res\\icon1.ico",
IMAGE_ICON, 0, 0, LR_LOADFROMFILE);
m_wndSBC.SetIcon(0, (HICON) hIcon);
But I'd still like to know why it didn't work the other way, if anyone can help.
Thanks.
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