Newbie question:
void main () {
int A = 1;
int B = 2;
double C = A / B;
stdout.printf("C value is: %g\n", C);
}
This prints: "C value is: 0"
void main () {
int A = 1;
double B = 2;
double C = A / B;
stdout.printf("C value is: %g\n", C);
}
This prints: "C value is: 0.5"
I don't understand the reason why the result is not 0.5 in both cases.
The division operation is performed on two integers, so the result is an integer. The fact that you assign it to a double afterwards doesn't change that.
What you're doing in your question, with the implicit conversions made explicit, is
int A = 1;
int B = 2;
double C = (double) (A / B);
However, if you want to perform the division operation using doubles you have to explicitly cast at least one of the operands to double:
int A = 1;
int B = 2;
double C = ((double) A) / B;
For the rules concerning arithmetic operations, see the arithmetic expressions section of the Vala Manaual. The relevant bit:
If both operands are of integer types, then the result will be the quotient only of the calculation (equivalent to the precise answer rounded down to an integer value.) If either operand is of a floating point type, then the result will be as precise as possible within the boundaries of the result type (which is worked out from the basic arithmetic type rules.)
Related
I have an Uint8List data list, for example:
Uint8List uintList = Uint8List.fromList([10, 1]);
How can I convert these numbers to a decimal number?
int decimalValue = ??? // in this case 265
Mees' answer is the correct general method, and it's good to understand how to do bitwise operations manually.
However, Dart does have a ByteData class that has various functions to help parse byte data for you (e.g. getInt16, getUint16). In your case, you can do:
Uint8List uintList = Uint8List.fromList([10, 1]);
int decimalValue = ByteData.view(uintList.buffer).getInt16(0, Endian.little);
print(decimalValue); // Prints: 266.
From what I understand of your question, you want decimalValue to be an integer where the least significant byte is (decimal)10, and the byte after that to be 1. This would result in the value 1 * 256 + 10 = 266. If you meant the bytes the other way around, it would be 10 * 256 + 1 = 2560 + 1 = 2561.
I don't actually have any experience with dart, but I assume code similar to this would work:
int decimalValue = 0;
for (int i = 0; i < uintList.length; i++) {
decimalValue = decimalValue << 8; // shift everything one byte to the left
decimalValue = decimalValue | uintList[i]; // bitwise or operation
}
If it doesn't produce the number you want it to, you might have to iterate through the loop backwards instead, which requires changing one line of code:
for (int i = uintList.length-1; i >= 0; i--) {
The compiler produce a "warning X3557: loop only executes for 0 iteration(s), forcing loop to unroll" and I don't understand why.
Here is the source code. It is a revisited itoa() function for HLSL producing resulting ascii codes in an array of uint.
#define ITOA_BUFFER_SIZE 16
// Convert uint to ascii and return number of characters
uint UIntToAscii(
in uint Num, // Number to convert
out uint Buf[ITOA_BUFFER_SIZE], // Where to put resulting ascii codes
in uint Base) // Numeration base for convertion
{
uint I, J, K;
I = 0;
while (I < (ITOA_BUFFER_SIZE - 1)) { // <==== Warning X3557
uint Digit = Num % Base;
if (Digit < 10)
Buf[I++] = '0' + Digit;
else
Buf[I++] = 'A' + Digit - 10;
if ((Num /= Base) == 0)
break;
}
// Reverse buffer
for (K = 0, J = I - 1; K < J; K++, J--) { // <==== Warning X3557
uint T = Buf[K];
Buf[K] = Buf[J];
Buf[J] = T;
}
// Fill remaining of buffer with zeros to make compiler happy
K = I;
while (K < ITOA_BUFFER_SIZE)
Buf[K++] = 0;
return I;
}
I tried to rewrite the while loop but this doesn't change anything. Also tried to use attribute [fastopt] without success. As far as I can see the function produce the correct result.
Any help appreciated.
The warning you are getting is
WAR_TOO_SIMPLE_LOOP 3557 The loop only executes for a limited number
of iterations or doesn't seem to do anything so consider removing it
or forcing it to unroll.
The warning is pretty much self explanatory, if you consider that loops are considered inefficient in GPGPU, so the compiler tries to unroll them when it's possible. What the compiler is telling you is that you created some loops that can run more efficiently if unrolled, or can be removed because they never run. If a loop is unrollable, it means that you can predict at compile time the number of times it will run. Your loops on first look should not fulfill this criterium.
I = 0;
while (I < (ITOA_BUFFER_SIZE - 1)) { // <==== Warning X3557
uint Digit = Num % Base;
if (Digit < 10)
Buf[I++] = '0' + Digit;
else
Buf[I++] = 'A' + Digit - 10;
if ((Num /= Base) == 0)
break;
}
This while loop runs up to 15 times I < (ITOA_BUFFER_SIZE - 1), depending on (Num /= Base) == 0. Final value of I is between 1 and 15, depending on how if ((Num /= Base) == 0) evaluates on each cycle. Nonetheless, it still is unrollable, because the compiler may still insert a conditional jump over the iterations.
// Reverse buffer
for (K = 0, J = I - 1; K < J; K++, J--) { // <==== Warning X3557
uint T = Buf[K];
Buf[K] = Buf[J];
Buf[J] = T;
}
This second loop, instead should not be unrollable, because I should not be known to the compiler.
The warning you reported
warning X3557: loop only executes for 0 iteration(s), forcing loop to unroll
might refer to the first loop if if ((Num /= Base) == 0) always evaluates to true on first iteration. In that case, I would be equal to 1, and J would be equal to 0 on the second loop. That second loop would not run because K < J would evaluate to false on first iteration.
What you get in the end if you let it [unroll] is probably a single iteration on the while loop and the complete removal of the subsequent for loop. I highly suspect this is not your intended behaviour, and while it might suppress the warning, you might want to check the code and see if something does not run the way it should.
I've prototyped an algorithm for my iOS game in Python, and I need to rewrite in in ObjC. Basically, I have a board of 16 numbers, and I want to loop through every number three times and the four functions I'm using (add, subtract, multiply, exponentiate). 1+2+3, 2*3-4, 3^4-5, 9-4^3, etc., but without order of operations (first operation is always done first).
What I would like is an overview of how this might be implemented in Objective-C. Specifically, what is the equivalent of an array of functions in Objective-C? Is there an easy way to implement it with selectors? What's the best structure to use for loops with numbers? Array of NSIntegers, array of ints, NSArray/NSMutableArray of NSNumbers?
import random as rand
min = 0
max = 9
max_target = 20
maximum_to_calculate = 100
def multiply(x, y):
return x * y
def exponate(x, y):
return x ** y
def add(x, y):
return x + y
def subtract(x, y):
return x - y
function_array = [multiply, exponate, add, subtract]
board = [rand.randint(min, max) for i in xrange(0, 16)]
dict_of_frequencies = {}
for a in board:
for b in board:
for first_fun in function_array:
first_result = first_fun(a, b)
for c in board:
for second_fun in function_array:
final_result = second_fun(first_result, c)
if final_result not in dict_of_frequencies:
dict_of_frequencies[final_result] = 0
dict_of_frequencies[final_result] += 1
The most convenient way in Objective-C to construct an array of functions would be to use Blocks:
typedef NSInteger (^ArithmeticBlock)(NSInteger, NSInteger);
ArithmeticBlock add = ^NSInteger (NSInteger x, NSInteger y){
return x + y;
};
ArithmeticBlock sub = ^NSInteger (NSInteger x, NSInteger y){
return x - y;
};
NSArray * operations = #[add, sub];
Since there's no great way to perform arithmetic on NSNumbers, it would probably be best to create and store the board's values as primitives, such as NSIntegers, in a plain C array. You can box them up later easily enough, if necessary -- #(boardValue) gives you an NSNumber.
If you want to do it with straight C function pointers, something like this will do it:
#include <stdio.h>
#include <math.h>
long add(int a, int b) {
return a + b;
}
long subtract(int a, int b) {
return a - b;
}
long multiply(int a, int b) {
return a * b;
}
long exponate(int a, int b) {
return pow(a, b);
}
int main(void) {
long (*mfunc[4])(int, int) = {add, subtract, multiply, exponate};
char ops[4] = {'+', '-', '*', '^'};
for ( int i = 0; i < 4; ++i ) {
printf("5 %c 9 = %ld\n", ops[i], mfunc[i](5, 9));
}
return 0;
}
and gives the output:
paul#MacBook:~/Documents/src$ ./rndfnc
5 + 9 = 14
5 - 9 = -4
5 * 9 = 45
5 ^ 9 = 1953125
paul#MacBook:~/Documents/src$
Function pointer syntax can be slightly convoluted. long (*mfunc[4])(int, int) basically translates to defining a four-element array, called mfunc, of pointers to functions returning long and taking two arguments of type int.
Maddy is right. Anyway, I'll give it a try just for the fun of it.
This has never seen a compiler. So please forgive me all the typos and minor syntax errors in advance.
#include <stdlib.h>
...
const int MIN = 0;
const int MAX = 9;
const int MAX_TARGET = 20;
const int MAX_TO_CALCULATE = 100;
...
- (int) multiply:(int)x with:(int)y { return x * y; }
- (int) exponate:(int)x with:(int)y { return x ^ y; }
- (int) add:(int)x to:(int)y { return x + y; }
- (int) substract:(int)x by:(int)y { return x - y; }
// some method should start here, probably with
-(void) someMethod {
NSArray *functionArray = [NSArray arrayWithObjects: #selector(multiply::), #selector(exponate::), #selector(add::), #substract(multiply::), nil]; // there are other ways of generating an array of objects
NSMutableArray *board = [NSMutableArray arrayWithCapacity:16]; //Again, there are other ways available.
for (int i = 0; i < 16; i++) {
[board addObject:#(arc4random() % (MAX-MIN) + MIN)];
}
NSMutableDictionary dictOfFrequencies = [[NSMutableDictionary alloc] init];
for (NSNumber a in board)
for (NSNumber b in board)
for (SEL firstFun in functionArray) {
NSNumber firstResult = #([self performSelector:firstFun withObject:a withObject:b]);
NSNumber countedResults = [dictOfFrequencies objectForKey:firstResult];
if (countedResults) {
[dictOfFrequencies removeObjectForKey:firstResult];
countedResults = #(1 + [countedResults intValue]);
} else {
countedResults = #1; // BTW, using the # followed by a numeric expression creates an NSNumber object with the value 1.
}
[dictOfFrequencies setObject:countedResults forKey:firstResult];
}
}
Well, let me add some comments before others do. :-)
There is no need for objective c. You python code is iterative therefore you can implement it in plain C. Plain C is available where ever Objective C is.
If you really want to go for Objective-C here then you should forget your python code and implement the same logic (aiming for the same result) in Objective-C in an OOP style. My code really tries to translate your code as close as possible. Therefore my code is far far away from neither beeing good style nor maintainable nor proper OOP. Just keep that in mind before you think, ObjC was complicated compared to python :-)
The following produces the below error:
int calc_ranks(ranks)
{
double multiplier = .5;
return multiplier * ranks;
}
The return type double is not a int, as defined by the method calc_ranks. How do I round/cast to an int?
Round it using the round() method:
int calc_ranks(ranks) {
double multiplier = .5;
return (multiplier * ranks).round();
}
You can use any of the following.
double d = 20.5;
int i = d.toInt(); // i = 20
int i = d.round(); // i = 21
int i = d.ceil(); // i = 21
int i = d.floor(); // i = 20
You can simply use toInt() to convert a num to an int.
int calc_ranks(ranks)
{
double multiplier = .5;
return (multiplier * ranks).toInt();
}
Note that to do exactly the same thing you can use the Truncating division operator :
int calc_ranks(ranks) => ranks ~/ 2;
I see a lot of answers, but with less description. Hope my answer will add some value.
Lets initalize the variable, and see how it will change with different methods.
double x = 8.5;
toInt()
It truncates the decimal value.
int a = x.toInt();
print(a); // 8
truncate()
It also truncates the decimal value.
int b = x.truncate();
print(b); // 8
round()
It returns the closest integer. It uses half up rounding mode.
int c = x.round();
print(c); // 9
ceil()
It returns the closest integer greater than the value.
int c = x.ceil();
print(c); // 9
floor()
It returns the closest integer smaller than the value.
int c = x.floor();
print(c); // 8
I looked at the answers above and added some more answers to make it a little easier to understand.
double value = 10.5;
Using toInt()
void main(){
double value = 10.5;
var y = value.toInt();
print(y);
print(y.runtimeType);
}
Using round()
The round() method returns the closest integer to the double.
void main(){
double value = 9.6;
var b = value.round();
print(b);
print(b.runtimeType);
}
Using ceil()
The ceil() method returns the smallest integer that is equal or greater than the given double.
void main(){
double value = 9.5;
var d = value.ceil();
print(d);
print(d.runtimeType);
}
Using floor()
The floor() method returns the greatest integer not greater than the given double.
void main(){
double value = 10.9;
var j = value.floor();
print(j);
print(j.runtimeType);
}
Conclusion
We’ve gone through 4 different techniques to convert a double to an integer in Dart and Flutter. You can choose from the method that fits your use case to solve your problem. Flutter is awesome and provides a lot of amazing features.
To convert double to int just this:
division
double01 ~/ double02
PS: The operator x ~/ y is more efficient than (x / y).toInt().
Multiplication, addition and subtraction
(double01 * double02).toInt
(double01 + double02).toInt
(double01 - double02).toInt
Its easy,
(20.8).round()
For String,
double.tryParse(20.8).round()
from string to int ->
if you string in int format like '10'
then use ---->
int.parse(value)
but if string in double format like '10.6'
then use like this ---->
double.parse(value).toInt()
convert double to int
doubleValue.toInt()
Try this!
int calc_ranks(ranks)
{
double multiplier = .5;
return (multiplier * ranks).truncate();
}
class CurrencyUtils{
static int doubletoint(double doublee) {
double multiplier = .5;
return (multiplier * doublee).round();
}
}
----------------------
CustomText( CurrencyUtils.doubletoint(
double.parse(projPageListss[0].budget.toString())
).toString(),
fontSize: 20,
color: Colors.white,
font: Font.QuicksandSemiBold,
),
There's another alternative, you can first cast the double to 'num' datatype and then convert to int using toInt().
double multiplier = .5;
return ((multiplier * ranks) as num).toInt();
The num type is an inherited data type of the int and double types.
You can cast both int and double to num, then cast it again to whatever you want
(double -> use toDouble(), int -> use toInt())
As much as I understand, I could parse the support vectors from the model produced by training with a set of data with LibSVM.
What would be the formula, to produce the classifier?
Do I need the data in the headers of the file, like the following (kernel etc...before the listed support vectors):
svm_type c_svc
kernel_type rbf
gamma 0.125
nr_class 4
total_sv 1038
rho -0.859244 -0.876628 -0.958343 0.543365 -1.10722 -1.79433
label 2 1 3 0
nr_sv 364 276 242 156
SV
My case is
I want to do classification from Node.JS. But there isn't any bindings for LibSVM for it, yet.
Since my models are not going to change, I would like to do the classification in Node.JS, holding the model in-memory.
If this proves to be slow, I rather write the same classification from scratch in C++ and create a wrapper module if it's only a matter of a simple computation (as I suspect it is).
Thanks.
You should be able to translate the C function to Javascript.
Here is the relevant code:
double svm_predict_values(const svm_model *model, const svm_node *x, double* dec_values)
{
int i;
int nr_class = model->nr_class;
int l = model->l;
double *kvalue = Malloc(double,l);
for(i=0;i<l;i++)
kvalue[i] = Kernel::k_function(x,model->SV[i],model->param);
int *start = Malloc(int,nr_class);
start[0] = 0;
for(i=1;i<nr_class;i++)
start[i] = start[i-1]+model->nSV[i-1];
int *vote = Malloc(int,nr_class);
for(i=0;i<nr_class;i++)
vote[i] = 0;
int p=0;
for(i=0;i<nr_class;i++)
for(int j=i+1;j<nr_class;j++)
{
double sum = 0;
int si = start[i];
int sj = start[j];
int ci = model->nSV[i];
int cj = model->nSV[j];
int k;
double *coef1 = model->sv_coef[j-1];
double *coef2 = model->sv_coef[i];
for(k=0;k<ci;k++)
sum += coef1[si+k] * kvalue[si+k];
for(k=0;k<cj;k++)
sum += coef2[sj+k] * kvalue[sj+k];
sum -= model->rho[p];
dec_values[p] = sum;
if(dec_values[p] > 0)
++vote[i];
else
++vote[j];
p++;
}
int vote_max_idx = 0;
for(i=1;i<nr_class;i++)
if(vote[i] > vote[vote_max_idx])
vote_max_idx = i;
free(kvalue);
free(start);
free(vote);
return model->label[vote_max_idx];
}
Notice that you have to recreate this equation:
The only difference is since your model has 4 classes, you need to implement the vote system which is basically the code above.
Hope it helps.