I would like to encode three keyboard modifiers (CTRL, ALT, SHIFT) + the ASCII code of the pressed key into a single value. This falls naturally into the category of bitmasks.
One way I could do this is that the sender encodes each key as the following:
CTRL: 1000
ALT: 10000
SHIFT: 100000
KeyCode: 1-255
For example, if I were to click all modifiers + the last key in the ascii table, I would get:
100000 + 10000 + 1000 + 255 = 111255. The receiver side it would then be possible to do substraction and check if the number goes below 0 as such:
has_shift = X - 100000 < 0
has_alt = X - 10000 < 0
has_ctrl = X - 1000 < 0
if has_shift
X -= 100000
if has_alt
X -= 10000
if has_ctrl
X -= 1000
keyCode = X (the remainder)
Surely enough, I find this horrible and would assume that this could be done in a far better using bit-shift or something in that ballpark. How could this possibly be done better?
Instead add 256, 512, and 1024 respectively for ctrl, alt, shift. Then use the and operator in whatever language you're using (missing from question tags) to extract the modifiers and code. In C and many languages, that operator is &. So X & 1024 is not zero if shift was pressed. X & 255 is the character code.
Related
I wrote a code that can get line projection (intensity profile) of an image, and I would like to convert/export this line projection (intensity profile) to excel table, and then order all the Y coordinate. For example, except the maximum and minimum values of all the Y coordinate, I would like to know largest 5 coordinate value and smallest coordinate value.
Is there any code can reach this function? Thanks,
image line_projection
Realimage imgexmp
imgexmp := GetFrontImage()
number samples = 256, xscale, yscale, xsize, ysize
GetSize( imgexmp, xsize, ysize )
line_projection := CreateFloatImage( "line projection", Xsize, 1 )
line_projection = 0
line_projection[icol,0] += imgexmp
line_projection /= samples
ShowImage( line_projection )
Finding a 'sorted' list of values
If you need to sort though large lists of values (i.e. large images) the following might not be very sufficient. However, if your aim is to get the "x highest" values with a relatively small number of X, then the following code is just fine:
number nFind = 10
image test := GetFrontImage().ImageClone()
Result( "\n\n" + nFind + " highest values:\n" )
number x,y,v
For( number i=0; i<nFind; i++ )
{
v = max(test,x,y)
Result( "\t" + v + " at " + x + "\n" )
test[x,y] = - Infinity()
}
Working with a copy and subsequently "removing" the maximum value by changing that pixel value. The max command is fast - even for large images -, but the for-loop iteration and setting of individual pixels is slow. Hence this script is too slow for a complete 'sorting' of the data if it is big, but it can quickly get you the n 'highest' values.
This is a non-coding answer:
If you havea LinePlot display in DigitalMicrograph, you can simply copy-paste that into Excel to get the numbers.
i.e. with the LinePlot image front most, preses CTRL + C to copy
(make sure there are no ROIs on it).
Switch to Excel and press CTRL + V. Done.
==>
i have this code in if statement , and i think it is very big , how can i change this by short code and for infinity i.e that i want the if statement to work on two and after two by sixteen i.e 2 += 16 , forever , i know that my question might be not understandable
but please help me
Take the current count and subtract 2. If the number is divisible by 16 (the remainder when you do modulo division is zero), then your statement is true.
if ((count - 2) % 16 == 0)
{
_secondBall.hidden = YES;
}
This can be reduced to a mathematical problem. What you want is to divide by 16 and see if there is no 'remainder'. The 'modulo' operator gives you this remainder. You can Google it if you want to understand. So, as others have quickly posted before me, modulo in Objective-C is done via %. So x % 16 means: divide x by 16 and return the remainder (or: subtract 16 from x until x is smaller than 16). x % 16 will be a number between 0 and 15. Always. When x % 16 is 0, it means x is dividable by 16.
Since you want to take action when count is 2 + 16 * n, you want to subtract 2 from count first. Like so if ( ( count - 2 ) % 16 == 0). Or you can do this, which is shorter but perhaps less easy to understand: if ( count % 16 == 2 ).
Try this:
if ((count - 2) % 16 == 0)
Use this syntax
if ((count-2) % 16 == 0)
How we can generate randomize number between a range in the Float numbers (in delphi xe3) ?
For example, randomize number between [0.10 to 0.90].
I need give results like:
[ 0.20 , 0.32 , 0.10 , 0.50 ]
Thanks for solutions....
Another option is to use RandomRange (returns: AFrom <= r < ATo) as follow:
RandomRange(10, 90 + 1) / 100
or
RandomRange(10, 90 + 1) * 0.01
will return numbers in the range of 0.10 to 0.90 (including 0.90)
var
float : Double;
float := Random; // Random float in range: 0 <= float < 1
float := 0.1 + float*0.8 // 0.1 <= float < 0.9
To initialize the Random number generator, make a single call to Randomizeor set the RandSeed parameter before calling the Random function for the first time.
Not doing so, generates the same sequence every time you run the program. Note however, that this sequence is not guaranteed when recompiling for another compiler version.
Try this:
function RandomRangeF(min, max: single): single;
begin
result := min + Random * (max - min);
end;
This is a bit cheeky but here goes: Depends how many numbers you want after the floating point. For example, if you want 1 number, you could generate in the 100 - 999 range and then divide by 10. Or 1000 - 9999 and divide by 100.
I'm trying the second day to send a midi signal. I'm using following code:
int pitchValue = 8191 //or -8192;
int msb = ?;
int lsb = ?;
UInt8 midiData[] = { 0xe0, msb, lsb};
[midi sendBytes:midiData size:sizeof(midiData)];
I don't understand how to calculate msb and lsb. I tried pitchValue << 8. But it's working incorrect, When I'm looking to events using midi tool I see min -8192 and +8064 max. I want to get -8192 and +8191.
Sorry if question is simple.
Pitch bend data is offset to avoid any sign bit concerns. The maximum negative deviation is sent as a value of zero, not -8192, so you have to compensate for that, something like this Python code:
def EncodePitchBend(value):
''' return a 2-tuple containing (msb, lsb) '''
if (value < -8192) or (value > 8191):
raise ValueError
value += 8192
return (((value >> 7) & 0x7F), (value & 0x7f))
Since MIDI data bytes are limited to 7 bits, you need to split pitchValue into two 7-bit values:
int msb = (pitchValue + 8192) >> 7 & 0x7F;
int lsb = (pitchValue + 8192) & 0x7F;
Edit: as #bgporter pointed out, pitch wheel values are offset by 8192 so that "zero" (i.e. the center position) is at 8192 (0x2000) so I edited my answer to offset pitchValue by 8192.
If I am given a floating point number but do not know beforehand what range the number will be in, is it possible to scale that number in some meaningful way to be in another range? I am thinking of checking to see if the number is in the range 0<=x<=1 and if not scale it to that range and then scale it to my final range. This previous post provides some good information, but it assumes the range of the original number is known beforehand.
You can't scale a number in a range if you don't know the range.
Maybe what you're looking for is the modulo operator. Modulo is basically the remainder of division, the operator in most languages is is %.
0 % 5 == 0
1 % 5 == 1
2 % 5 == 2
3 % 5 == 3
4 % 5 == 4
5 % 5 == 0
6 % 5 == 1
7 % 5 == 2
...
Sure it is not possible. You can define range and ignore all extrinsic values. Or, you can collect statistics to find range in run time (i.e. via histogram analysis).
Is it really about image processing? There are lots of related problems in image segmentation field.
You want to scale a single random floating point number to be between 0 and 1, but you don't know the range of the number?
What should 99.001 be scaled to? If the range of the random number was [99, 100], then our scaled-number should be pretty close to 0. If the range of the random number was [0, 100], then our scaled-number should be pretty close to 1.
In the real world, you always have some sort of information about the range (either the range itself, or how wide it is). Without further info, the answer is "No, it can't be done."
I think the best you can do is something like this:
int scale(x) {
if (x < -1) return 1 / x - 2;
if (x > 1) return 2 - 1 / x;
return x;
}
This function is monotonic, and has a range of -2 to 2, but it's not strictly a scaling.
I am assuming that you have the result of some 2-dimensional measurements and want to display them in color or grayscale. For that, I would first want to find the maximum and minimum and then scale between these two values.
static double[][] scale(double[][] in, double outMin, double outMax) {
double inMin = Double.POSITIVE_INFINITY;
double inMax = Double.NEGATIVE_INFINITY;
for (double[] inRow : in) {
for (double d : inRow) {
if (d < inMin)
inMin = d;
if (d > inMax)
inMax = d;
}
}
double inRange = inMax - inMin;
double outRange = outMax - outMin;
double[][] out = new double[in.length][in[0].length];
for (double[] inRow : in) {
double[] outRow = new double[inRow.length];
for (int j = 0; j < inRow.length; j++) {
double normalized = (inRow[j] - inMin) / inRange; // 0 .. 1
outRow[j] = outMin + normalized * outRange;
}
}
return out;
}
This code is untested and just shows the general idea. It further assumes that all your input data is in a "reasonable" range, away from infinity and NaN.