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First call GetRichEditCtrl() to get a reference (not a pointer) to the view's edit control, then call SetWindowText(), like:
CRichEditCtrl &edit = GetRichEditCtrl();
edit.SetWindowText("Hello");
"The largest fire starts but with the smallest spark." - David Crow
"Judge not by the eye but by the heart." - Native American Proverb
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 I must have posted at the same time as you did.
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The following obtains a pointer to the view's control the outputs a string to the current caret position. For example:
void MyRichEdView::WriteString(CString sStr)
{
CRichEditCtrl & rich = GetRichEditCtrl();
rich.ReplaceSel(sStr);
}
Mark
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Hello,
I am developing an application which must acquire films with the V700 scanner using TWAIN; i have problems defining the scan area; with the older Expression 1680pro I have no problems at all. Has anyone experience with this scanner? Help is VERY appreciated...
Regards, Fulvio
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I am using C++ Builder Pro to create a database program. I have a DBCtrlGrid which views several records at one time. On this grid I have various text boxes with the data from the database records displayed. I would like to write a statement where if a paticular value is > 10 the text would change to red for that specific record textbox(123Txt). when I try to do this now it changes all the viewed records text in the textbox(123Txt) to red. Is it possible to change just one panels txt using this DBCtrlGrid? Any thoughts would be great.
Debs
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I need to take a float or double value nand determine how many numbers it has before and after the decimal. Examples:
123.3 -> 3 before and 1 after
.12345 -> 0 before and 5 after
123 -> 3 before and 0 after
The method has to be extremely fast because it may be executed very often in a high-volume environment. I know I can convert it to a string and search for the decimal, etc., but string functions are very slow.
Thanks!
Barry Etter
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use CString's Format function
CString strFloat;
strFloat.Format("%f",myfloatvalue);
//Search for .
strFloat.Find();
and then use CString's Left() and Right() methods
This code is not optimized. It's just one way of doing this
Somethings seem HARD to do, until we know how to do them.
_AnShUmAn_
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Yeap buddy,
I read that , That's why I quoted that the method I propose is not optimized and there could be many other methods of achieving the same.
Somethings seem HARD to do, until we know how to do them.
_AnShUmAn_
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Do you know how how a double or float laid in memory?
it would be easier using some mask & ing, | ing or <<ing>>ing to catch the values.
SaRath.
"It is your attitude, not your aptitude, that determines your altitude - Zig Ziglar."
My Blog | Understanding State Pattern in C++
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He specifically said that he did not want to convert to and from strings.
--
100% natural. No superstitious additives.
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For the number of digits before the decimal, you could use
int iNumBefore = 0;<br />
iNumBefore = 1 + (int)(log10(x));
where x is your value
As for the number of digits after, I don't have any quick ideas on that, but as you say, you could convert to a string (although you might get different results because of the precision of doubles) and then find the decimal.
Hope that helps.
Karl - WK5M
PP-ASEL-IA (N43CS)
<kmedcalf@ev1.net>
PGP Key: 0xDB02E193
PGP Key Fingerprint: 8F06 5A2E 2735 892B 821C 871A 0411 94EA DB02 E193
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krmed wrote: As for the number of digits after, I don't have any quick ideas on that, but as you say, you could convert to a string (although you might get different results because of the precision of doubles) and then find the decimal.
<br />
int iNumAfter = 0;<br />
iNumAfter = 1 + (int)(log10(x*10));<br />
Could it works?
F.
TBEditor: a pandapowered app!
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To find the number of digits before the floating point, you can use the log10 function. You need to round down the result and add 1.
For the number of digits after the floating point, I don't have any idea.
Cédric Moonen
Software developer
Charting control
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Cedric Moonen wrote: For the number of digits after the floating point, I don't have any idea.
well you can do the same... just multiply the initial number by 10 ...
F.
TBEditor: a pandapowered app!
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I think you can investigate the following approach.
For the "before" part of positive values:
float f;
long m = (long)f;
int before;
if( m == 0) before = 0;
else if( m <= 9) before = 1;
else if( m <= 99) before = 2;
else if( m <= 999) before = 3;
else ... and so on according to a maximal value.
Hope it is fast enough.
For the "after" part, you can try a similar idea, but I am afraid it will be some problems related to imprecise representation of decimal float numbers in binary format, but such issue you will see in case of string-convertion-based method too.
-- modified at 10:35 Wednesday 28th June, 2006
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Sarath.<fon wrote:<="" div="">can't we use mod (%) instead?
I suppose it is not faster.
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double x(123.3)
ceil(log10(x)) => 3 ie the # digits left of decimal.
I'm still working of # digits right of decimal...
Regards,
Kylur
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Be careful of the behaviour if x = 0 when using "log10".
For the number of digits after "." I think you're going to have some problems, simply because the computer representation of floats and doubles of numbers do not exactly correspond to the value you gave. For instance it could happen that 123.3 actually corresponds to 123.29999999999999 as soon as it is stored as a float. So trying to guess the number of digits after "." makes no sense. You could maybe get around this if you limit the possible amount of decimal numbers to a well chosen value.
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you're going to run into floating-point precision errors, unless you round numbers to a maximum number of decimal places. ex. you're going to come across something like 1.9999999999...
Cleek | Image Toolkits | Thumbnail maker
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String operations are your best bet for the most accuracy and flexibility (easy to control the number of significant figures). Since you don't want to go that route, this might get you started:
To count left side of decimal
<br />
unsigned long countLeftOfDecimal(double x)<br />
{<br />
if (1 > abs(x))<br />
{<br />
return 0;
}<br />
else if (1 == abs(x))<br />
{<br />
return 1;
}<br />
else<br />
{<br />
return ceil(log10(abs(x)));<br />
}<br />
}<br />
<br />
unsigned long countRightOfDecimal(double x)<br />
{<br />
long y = x;
double x1 = x - (double)y;
long count = 0;<br />
while (0.0 != x1)<br />
{<br />
double x2 = x1 * 10.0;
y = x2;
x1 = x2 - y;
++count;<br />
}<br />
return count;<br />
}<br />
Keep in mind that you will run into problems when you have numbers like 1.99999999999... with this algorithm -- which is why converting to a string for counting significant figures is a better solution, and may actually be faster than any non-string solution you come up with that doesn't have problems with infinite repeating decimals.
If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week
Zac
-- modified at 15:16 Wednesday 28th June, 2006
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I don't think there is an easy solution for this without passing the real values to a string. This is mainly due to the fact that the real numbers may have an exact representation as decimal and an inexact representation as native. And, of course, vice-versa.
For example, float 0.2 has an inexact native representation. The exact representation is 0.200000002980232. But the resolution present in a float is not enough for all the decimals.
This just means that you probably won't have much luch analysing the bits in the float (or double). Their memory layout can be found in http://www.psc.edu/general/software/packages/ieee/ieee.html[^].
In any case, finding the values you want by using strings or not, this can always get complicated:
float f=0.00031233478376329384729344178300483284723740923762730534976862e7f;
double d=f;
printf("f=%f, d=%.15g", f, d);
This code compiles, but due to the lack of precision in the floating point representations the results are not what one would expect.
So, the only fast solution I can imagine involves the assumption that you have not that many number variations, they are always around zero (controlled exponent), and that you do not have rounding errors. With these assumptions you could build a fast look up table.
For example, assume that you will have numbers between -999.99 and +999.99. This results in 199998 diferent numbers plus the zero, i.e., 199999. So you could program the following algorithm:
// build table
float f;
int i;
i=0;
for(f=-999.99; f<=-999.99; f+=0.01f) {
table[i].left=ComputeLeftByUsingStrings(f);
table[i].right=ComputeRightByUsingStrings(f);
i++;
}
// get the left/right values for float f fast!
f*=100.0f; // shift so that the smallest decimal becomes an integer
int i=(int)f; // convert to int to use as offset to the table
i+=99999; // correct the start offset to start at 0 for the most negative number
int left=table[i].left; // left will contain the number of characters to the left
int right=table[i].right; // right will contain the number of characters to the right
I hope this helps!
Rilhas
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