bitmask enumerations in objective-c or C - iphone

I'm trying to draw a line, broken out into segments dependent on values. For example, if there are 5 fields, and all 5 fields were true, then my Line would look like
-----
If say the first and last fields were true, and everything else would be false, then it would be
- -
I thought I could do this with a bit mask of some sort. First of all, I've never done a bit mask before, but I think I've seen them here and there. I was wondering how I could go about this, and use enumerations instead of 1/0 for readability. As far as I can see from my data, I would only need values of either 1 or 0 for the different properties. However, it would be good to know how to have one of the values be a three level or higher enumeration for future reference. Thanks!
Trying to do something like:
enum CodingRegions {
Coding = 0x01,
NonCoding = 0x02
};
enum Substitution {
Synonymous = 0x04,
NonSynonmous = 0x05
};
Then based on the value of the object, I could do
bitmask???? = object.CodingRegion | object.Substitution;
Then later, check the value of the bitmask somehow, and then draw the line accordingly based on what the values are.

Not sure exactly what your requirements are, but here is one way it might be written in C:
#include <stdio.h>
typedef enum MyField_ {
hasWombat = 1 << 0,
hasTrinket = 1 << 1,
hasTinRoof = 1 << 2,
hasThreeWheels = 1 << 3,
myFieldEnd = 1 << 4,
} MyField;
void printMyField(MyField data) {
MyField field = 1;
while (field != myFieldEnd) {
printf("%c", data & field ? '-' : ' ');
field <<= 1;
}
printf("\n");
}
int main() {
MyField data = hasTrinket | hasThreeWheels;
printMyField(data);
data |= hasWombat; // set a field
data &= ~hasTrinket; // clear a field
printMyField(data);
return 0;
}

Not sure this is what you want, but:
// assumed Coding/NonCoding, Synonomous/NonSynonymous are opposites of each other. If not, add more bit fields
enum CodingRegions
{
Coding = 1 << 0
} ;
enum Substitution
{
Synonymous = 1 << 1
}
void PrintBitmask( NSUInteger bitmask )
{
printf( "%s", bitmask & Coding != 0 ? "-" : " " ) ;
printf( "%s", bitmask & Substitution != 0 ? "-" : " " ) ;
printf( "\n" ) ;
}
Your PrintBitmask() could also look like this:
void PrintBitmask( NSUInteger bitmask )
{
printf( "%s", bitmask & Coding != 0 ? "Coding" : "Noncoding" ) ;
printf( "|" ) ;
printf( "%s", bitmask & Substitution != 0 ? "Synonymous-" : "Nonsynonymous" ) ;
printf( "\n" ) ;
}

/* I prefer macros over enums (at least for something this simple) */
#define SPACE 0x0
#define DASH 0x1
/* input fields */
int fields[5] = {DASH,SPACE,SPACE,SPACE,DASH};
/* create bitmask */
for (int i=0; i<5; i++) {
mask |= (fields[i] << i);
}
/* interpret bitmask and print the line */
for (int i=0; i<5; i++) {
if (mask & (fields[i] << i)) {
printf("%c", '-');
} else {
printf("%c", ' ');
}
}

Related

Type conversion - string of characters to integer

Hello I am writing my program in C, using PSoC tools to program my Cypress development kit. I am facing an issue regarding type conversion of a string of characters collected in my circular buffer (buffer) to a local variable "input_R", ultimately to a global variable st_input_R. The event in my FSM calling this action function is given below:
void st_state_5_event_0(void) //S6 OR S4
{
char buffer[ST_NODE_LIMIT] = {0};
st_copy_buffer(buffer);
uint32 input_R = {0};
mi_utoa(input_R, buffer);
if ((input_R >= 19000) && (input_R <= 26000))
{
st_input_R = input_R;
_st_data.state = ST_STATE_6;
}
else
{
_st_data.status = ST_STATE_4;
}
UART_1_Stop();
st_stop();
st_empty_buffer();
}
ST_NODE_LIMIT = 64
st_copy_buffer copies the the numbers I type in using hyper terminal to the circular buffer named "buffer".
input_R is the 32 bit integer I want the buffer content to be converted to.
mi_utoa is the function I am using for converting the contents in the buffer to input_R and is detailed below:
uint8 mi_utoa(uint32 number, char *string)
{
uint8 result = MI_BAD_ARGUMENT;
if (string != NULL)
{
uint8 c = 0;
uint8 i = 0;
uint8 j = 0;
do
{
string[i++] = number % 10 + '0';
} while ((number /=10) > 0);
string[i] = '\0';
for (i = 0, j = strlen(string) - 1 ; i < j ; i++, j--)
{
c = string[i];
string[i] = string[j];
string[j] = c;
}
result = MI_SUCCESS;
}
return result;
}
The problem is, suppose if I enter 21500(+\r), the mi_utoa function converts the first digit to 0 the second digit to \000 while the other digits including the carriage return "\r" remains unaltered. As a result the input_R is NOT = 21500. Its happening for any string of digits I input. So the condition "if ((input_R >= 19000) && (input_R <= 26000))" is never satisfied. Hence the FSM returns to state 4 all the time and I am going in circles.
Can you please advice where the bug is in the mi_utoa function? Let me know if you want to know any other details.
Your function st_state_5_event_0() sets the value input_R to zero. Then you call mi_utoa(), which converts the value input_R to an ascii string, "0".
void st_state_5_event_0(void) //S6 OR S4
{
char buffer[ST_NODE_LIMIT] = {0};
//what is the value of buffer after this statement?
st_copy_buffer(buffer);
//the value of input_R after the next statement is =0
uint32 input_R = {0};
//conversion of input_R to string will give ="0"
mi_utoa(input_R, buffer);
if ((input_R >= 19000) && (input_R <= 26000))
{
st_input_R = input_R;
_st_data.state = ST_STATE_6;
}
//...
}
You probably want a function which converts your ascii buffer to a number.
uint8
mi_atou(uint32* number, char *string)
{
uint8 result = MI_BAD_ARGUMENT;
if (!string) return result;
if (!number) return result;
uint8 ndx = 0;
uint32 accum=0;
for( ndx=0; string[ndx]; ++ndx )
{
if( (string[ndx] >= '0') && (string[ndx] <= '9') )
{
accum = accum*10 + (string[ndx]-'0');
//printf("[%d] %s -> %d\n",ndx,string,accum);
}
else break;
}
//printf("[%d] %s -> %d\n",ndx,string,accum);
*number = accum;
result = MI_SUCCESS;
return result;
}
Which you would call by providing the address of the number to store the result,
mi_atou(&input_R, buffer);

does kyoto cabinet support key range search?

Does Kyoto Cabinet support searching for a range of keys?
If so, what types of keys do support range search?
Can I do range search on a long (64bit) key?
Thanks
RG
it supports key prefix query, however, the efficiency of prefix query depends on what internal storage structure is. If you are using hashdb, it may be not a good idea, as keys & values are scattered around in the underline file.
Yes, for integers.
B+ tree database supports sequential access in order of the keys, which realizes forward matching search for strings and range search for integers - from docs
Yes you can, you just need a forward jump.
An example using C. Stores 5 records with 64 bits keys (from 1 to 5) and then apply a filter (from 2 to 4):
#include <kclangc.h>
#include <inttypes.h>
int main(void)
{
KCDB *db;
KCCUR *cur;
char *kbuf;
size_t ksiz, vsiz;
const char *cvbuf;
int64_t i, val, min, max;
int64_t keys[] = {1, 2, 3, 4, 5};
const char *values[] = {"one", "two", "three", "four", "five"};
char i64[8]; /* A buffer to store byte sequences */
/* create the database object */
db = kcdbnew();
/* open the database */
if (!kcdbopen(db, "db64.kct", KCOWRITER | KCOCREATE)) {
fprintf(stderr, "open error: %s\n", kcecodename(kcdbecode(db)));
}
/* store records */
for (i = 0; i < 5; i++) {
memcpy(i64, &keys[i], 8);
if (!kcdbset(db, i64, 8, values[i], strlen(values[i]))) {
fprintf(stderr, "set error: %s\n", kcecodename(kcdbecode(db)));
exit(EXIT_FAILURE);
}
}
/* traverse records */
min = 2;
max = 4;
printf("Range from %" PRId64 " to %" PRId64 "\n", min, max);
memcpy(i64, &min, 8);
cur = kcdbcursor(db);
kccurjumpkey(cur, i64, 8);
while ((kbuf = kccurget(cur, &ksiz, &cvbuf, &vsiz, 1)) != NULL) {
memcpy(&val, kbuf, 8);
if (val > max) {
break;
}
printf("Found %s\n", cvbuf);
kcfree(kbuf);
}
kccurdel(cur);
/* close the database */
if (!kcdbclose(db)) {
fprintf(stderr, "close error: %s\n", kcecodename(kcdbecode(db)));
}
/* delete the database object */
kcdbdel(db);
return 0;
}
LevelDB supports binary keys and ranged queries.
Edit: I forgot to mention that in order for the range query to work, the binary value needs to be packed in a comparable way. For your long example, you need to make sure it's big-endian encoded.

Carefully deleting N items from a "circular" vector (or perhaps just an NSMutableArray)

Imagine a std:vector, say, with 100 things on it (0 to 99) currently. You are treating it as a loop. So the 105th item is index 4; forward 7 from index 98 is 5.
You want to delete N items after index position P.
So, delete 5 items after index 50; easy.
Or 5 items after index 99: as you delete 0 five times, or 4 through 0, noting that position at 99 will be erased from existence.
Worst, 5 items after index 97 - you have to deal with both modes of deletion.
What's the elegant and solid approach?
Here's a boring routine I wrote
-(void)knotRemovalHelper:(NSMutableArray*)original
after:(NSInteger)nn howManyToDelete:(NSInteger)desired
{
#define ORCO ((NSInteger)[original count])
static NSInteger kount, howManyUntilLoop, howManyExtraAferLoop;
if ( ... our array is NOT a loop ... )
// trivial, if messy...
{
for ( kount = 1; kount<=desired; ++kount )
{
if ( (nn+1) >= ORCO )
return;
[original removeObjectAtIndex:( nn+1 )];
}
return;
}
else // our array is a loop
// messy, confusing and inelegant. how to improve?
// here we go...
{
howManyUntilLoop = (ORCO-1) - nn;
if ( howManyUntilLoop > desired )
{
for ( kount = 1; kount<=desired; ++kount )
[original removeObjectAtIndex:( nn+1 )];
return;
}
howManyExtraAferLoop = desired - howManyUntilLoop;
for ( kount = 1; kount<=howManyUntilLoop; ++kount )
[original removeObjectAtIndex:( nn+1 )];
for ( kount = 1; kount<=howManyExtraAferLoop; ++kount )
[original removeObjectAtIndex:0];
return;
}
#undef ORCO
}
Update!
InVariant's second answer leads to the following excellent solution. "starting with" is much better than "starting after". So the routine now uses "start with". Invariant's second answer leads to this very simple solution...
N times do if P < currentsize remove P else remove 0
-(void)removeLoopilyFrom:(NSMutableArray*)ra
startingWithThisOne:(NSInteger)removeThisOneFirst
howManyToDelete:(NSInteger)countToDelete
{
// exception if removeThisOneFirst > ra highestIndex
// exception if countToDelete is > ra size
// so easy thanks to Invariant:
for ( do this countToDelete times )
{
if ( removeThisOneFirst < [ra count] )
[ra removeObjectAtIndex:removeThisOneFirst];
else
[ra removeObjectAtIndex:0];
}
}
Update!
Toolbox has pointed out the excellent idea of working to a new array - super KISS.
Here's an idea off the top of my head.
First, generate an array of integers representing the indices to remove. So "remove 5 from index 97" would generate [97,98,99,0,1]. This can be done with the application of a simple modulus operator.
Then, sort this array descending giving [99,98,97,1,0] and then remove the entries in that order.
Should work in all cases.
This solution seems to work, and it copies all remaining elements in the vector only once (to their final destination).
Assume kNumElements, kStartIndex, and kNumToRemove are defined as const size_t values.
vector<int> my_vec(kNumElements);
for (size_t i = 0; i < my_vec.size(); ++i) {
my_vec[i] = i;
}
for (size_t i = 0, cur = 0; i < my_vec.size(); ++i) {
// What is the "distance" from the current index to the start, taking
// into account the wrapping behavior?
size_t distance = (i + kNumElements - kStartIndex) % kNumElements;
// If it's not one of the ones to remove, then we keep it by copying it
// into its proper place.
if (distance >= kNumToRemove) {
my_vec[cur++] = my_vec[i];
}
}
my_vec.resize(kNumElements - kNumToRemove);
There's nothing wrong with two loop solutions as long as they're readable and don't do anything redundant. I don't know Objective-C syntax, but here's the pseudocode approach I'd take:
endIdx = after + howManyToDelete
if (Len <= after + howManyToDelete) //will have a second loop
firstloop = Len - after; //handle end in the first loop, beginning in second
else
firstpass = howManyToDelete; //the first loop will get them all
for (kount = 0; kount < firstpass; kount++)
remove after+1
for ( ; kount < howManyToDelete; kount++) //if firstpass < howManyToDelete, clean up leftovers
remove 0
This solution doesn't use mod, does the limit calculation outside the loop, and touches the relevant samples once each. The second for loop won't execute if all the samples were handled in the first loop.
The common way to do this in DSP is with a circular buffer. This is just a fixed length buffer with two associated counters:
//make sure BUFSIZE is a power of 2 for quick mod trick
#define BUFSIZE 1024
int CircBuf[BUFSIZE];
int InCtr, OutCtr;
void PutData(int *Buf, int count) {
int srcCtr;
int destCtr = InCtr & (BUFSIZE - 1); // if BUFSIZE is a power of 2, equivalent to and faster than destCtr = InCtr % BUFSIZE
for (srcCtr = 0; (srcCtr < count) && (destCtr < BUFSIZE); srcCtr++, destCtr++)
CircBuf[destCtr] = Buf[srcCtr];
for (destCtr = 0; srcCtr < count; srcCtr++, destCtr++)
CircBuf[destCtr] = Buf[srcCtr];
InCtr += count;
}
void GetData(int *Buf, int count) {
int srcCtr = OutCtr & (BUFSIZE - 1);
int destCtr = 0;
for (destCtr = 0; (srcCtr < BUFSIZE) && (destCtr < count); srcCtr++, destCtr++)
Buf[destCtr] = CircBuf[srcCtr];
for (srcCtr = 0; srcCtr < count; srcCtr++, destCtr++)
Buf[destCtr] = CircBuf[srcCtr];
OutCtr += count;
}
int BufferOverflow() {
return ((InCtr - OutCtr) > BUFSIZE);
}
This is pretty lightweight, but effective. And aside from the ctr = BigCtr & (SIZE-1) stuff, I'd argue it's highly readable. The only reason for the & trick is in old DSP environments, mod was an expensive operation so for something that ran often, like every time a buffer was ready for processing, you'd find ways to remove stuff like that. And if you were doing FFT's, your buffers were probably a power of 2 anyway.
These days, of course, you have 1 GHz processors and magically resizing arrays. You kids get off my lawn.
Another method:
N times do {remove entry at index P mod max(ArraySize, P)}
Example:
N=5, P=97, ArraySize=100
1: max(100, 97)=100 so remove at 97%100 = 97
2: max(99, 97)=99 so remove at 97%99 = 97 // array size is now 99
3: max(98, 97)=98 so remove at 97%98 = 97
4: max(97, 97)=97 so remove at 97%97 = 0
5: max(96, 97)=97 so remove at 97%97 = 0
I don't program iphone for know, so I image std::vector, it's quite easy, simple and elegant enough:
#include <iostream>
using std::cout;
#include <vector>
using std::vector;
#include <cassert> //no need for using, assert is macro
template<typename T>
void eraseCircularVector(vector<T> & vec, size_t position, size_t count)
{
assert(count <= vec.size());
if (count > 0)
{
position %= vec.size(); //normalize position
size_t positionEnd = (position + count) % vec.size();
if (positionEnd < position)
{
vec.erase(vec.begin() + position, vec.end());
vec.erase(vec.begin(), vec.begin() + positionEnd);
}
else
vec.erase(vec.begin() + position, vec.begin() + positionEnd);
}
}
int main()
{
vector<int> values;
for (int i = 0; i < 10; ++i)
values.push_back(i);
cout << "Values: ";
for (vector<int>::const_iterator cit = values.begin(); cit != values.end(); cit++)
cout << *cit << ' ';
cout << '\n';
eraseCircularVector(values, 5, 1); //remains 9: 0,1,2,3,4,6,7,8,9
eraseCircularVector(values, 16, 5); //remains 4: 3,4,6,7
cout << "Values: ";
for (vector<int>::const_iterator cit = values.begin(); cit != values.end(); cit++)
cout << *cit << ' ';
cout << '\n';
return 0;
}
However, you might consider:
creating new loop_vector class, if you use this kind of functionality enough
using list if you perform many deletions (or few deletions (not from end, that's simple pop_back) but large array)
If your container (NSMutableArray or whatever) is not list, but vector (i.e. resizable array), you most definitely don't want to delete items one by one, but whole range (e.g. std::vector's erase(begin, end)!
Edit: reacting to comment, to fully realize what must be done by vector, if you erase element other than the last one: it must copy all values after that element (e.g. 1000 items in array, you erase first, 999x copying (moving) of item, that is very costly).
Example:
#include <iostream>
#include <vector>
#include <ctime>
using namespace std;
int main()
{
clock_t start, end;
vector<int> vec;
const int items = 64 * 1024;
cout << "using " << items << " items in vector\n";
for (size_t i = 0; i < items; ++i) vec.push_back(i);
start = clock();
while (!vec.empty()) vec.erase(vec.begin());
end = clock();
cout << "Inefficient method took: "
<< (end - start) * 1.0 / CLOCKS_PER_SEC << " ms\n";
for (size_t i = 0; i < items; ++i) vec.push_back(i);
start = clock();
vec.erase(vec.begin(), vec.end());
end = clock();
cout << "Efficient method took: "
<< (end - start) * 1.0 / CLOCKS_PER_SEC << " ms\n";
return 0;
}
Produces output:
using 65536 items in vector
Inefficient method took: 1.705 ms
Efficient method took: 0 ms
Note it's very easy to get inefficient, look e.g. have at http://www.cplusplus.com/reference/stl/vector/erase/

converting Biginteger to Bytearray(Raw data)

I have used the following code for converting the bigint in decimal to bytearray (raw data), but I'm getting wrong result.
What is the mistake here?
I'm trying this in Apple Mac ( for Iphone app)
COMP_BYTE_SIZE is 4
Is there any bigendian/ little endian issue, please Help.
void bi_export(BI_CTX *ctx, bigint *x, uint8_t *data, int size)
{
int i, j, k = size-1;
check(x);
memset(data, 0, size); /* ensure all leading 0's are cleared */
for (i = 0; i < x->size; i++)
{
for (j = 0; j < COMP_BYTE_SIZE; j++)
{
comp mask = 0xff << (j*8);
int num = (x->comps[i] & mask) >> (j*8);
data[k--] = num;
if (k < 0)
{
break;
}
}
}
Thanks.
The argument size is at least x->size*4, ie. the target array is big enough? Also use
comp mask = (comp)0xff << (j*8);
num should be cast to uint8_t before copy
data[k--] = (uint8_t)num;

Formatting a (large) number "12345" to "12,345"

Say I have a large number (integer or float) like 12345 and I want it to look like 12,345.
How would I accomplish that?
I'm trying to do this for an iPhone app, so something in Objective-C or C would be nice.
Here is the answer.
NSNumber* number = [NSNumber numberWithDouble:10000000];
NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
[numberFormatter setNumberStyle:kCFNumberFormatterDecimalStyle];
[numberFormatter setGroupingSeparator:#","];
NSString* commaString = [numberFormatter stringForObjectValue:number];
[numberFormatter release];
NSLog(#"%# -> %#", number, commaString);
Try using an NSNumberFormatter.
This should allow you to handle this correctly on an iPhone. Make sure you use the 10.4+ style, though. From that page:
"iPhone OS: The v10.0 compatibility mode is not available on iPhone OS—only the 10.4 mode is available."
At least on Mac OS X, you can just use the "'" string formatter with printf(3).
$ man 3 printf
`'' Decimal conversions (d, u, or i) or the integral portion
of a floating point conversion (f or F) should be
grouped and separated by thousands using the non-mone-
tary separator returned by localeconv(3).
as in printf("%'6d",1000000);
Cleaner C code
// write integer value in ASCII into buf of size bufSize, inserting commas at tousands
// character string in buf is terminated by 0.
// return length of character string or bufSize+1 if buf is too small.
size_t int2str( char *buf, size_t bufSize, int val )
{
char *p;
size_t len, neg;
// handle easy case of value 0 first
if( val == 0 )
{
a[0] = '0';
a[1] = '\0';
return 1;
}
// extract sign of value and set val to absolute value
if( val < 0 )
{
val = -val;
neg = 1;
}
else
neg = 0;
// initialize encoding
p = buf + bufSize;
*--p = '\0';
len = 1;
// while the buffer is not yet full
while( len < bufSize )
{
// put front next digit
*--p = '0' + val % 10;
val /= 10;
++len;
// if the value has become 0 we are done
if( val == 0 )
break;
// increment length and if it's a multiple of 3 put front a comma
if( (len % 3) == 0 )
*--p = ',';
}
// if buffer is too small return bufSize +1
if( len == bufSize && (val > 0 || neg == 1) )
return bufSize + 1;
// add negative sign if required
if( neg == 1 )
{
*--p = '-';
++len;
}
// move string to front of buffer if required
if( p != buf )
while( *buf++ = *p++ );
// return encoded string length not including \0
return len-1;
}
I did this for an iPhone game recently. I was using the built-in LCD font, which is a monospaced font. I formatted the numbers, ignoring the commas, then stuck the commas in afterward. (The way calculators do it, where the comma is not considered a character.)
Check out the screenshots at RetroJuJu. Sorry--they aren't full-sized screenshots so you'll have to squint!
Hope that helps you (it's in C) :
char* intToFormat(int a)
{
int nb = 0;
int i = 1;
char* res;
res = (char*)malloc(12*sizeof(char));
// Should be enough to get you in the billions. Get it higher if you need
// to use bigger numbers.
while(a > 0)
{
if( nb > 3 && nb%3 == 0)
res[nb++] = ',';
// Get the code for the '0' char and add it the position of the
// number to add (ex: '0' + 5 = '5')
res[nb] = '0' + a%10;
nb++;
a /= 10;
}
reverse(&res);
return res;
}
There might be a few errors I didn't see (I'm blind when it comes to this...)
It's like an enhanced iToA so maybe it's not the best solution.
Use recursion, Luke:
#include <stdio.h>
#include <stdlib.h>
static int sprint64u( char* buffer, unsigned __int64 x) {
unsigned __int64 quot = x / 1000;
int chars_written;
if ( quot != 0) {
chars_written = sprint64u( buffer, quot);
chars_written += sprintf( buffer + chars_written, ".%03u", ( unsigned int)( x % 1000));
}
else {
chars_written = sprintf( buffer, "%u", ( unsigned int)( x % 1000));
}
return chars_written;
}
int main( void) {
char buffer[ 32];
sprint64u( buffer, 0x100000000ULL);
puts( buffer);
return EXIT_SUCCESS;
}